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

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

122 PAÑCASIDDHĀNTIKĀ IV. 49 whether actually the constants are 6 and 12, as here or 5 and 10, or some other number and double that, depends upon the accuracy of the result we get. For the matter of that there is another rule, very popular and attributed to our author himself in the following form: time = 5 × daytime ÷ (shadow − mid-day shadow + 10), given by the popular verse: chāyā nijeṣṭā dinamadhyabhāgacchāyonitā diksahitā tayāpte | dine śaraghne gatagamyanāḍīḥ śrīmān Varāho vadati syayuktyā || Here too the shadow is that of the 12 digit gnomon. Note that the multiplier here is 5, and the additive constant double that, viz. 10. Actually, different constants for different places, and for different times, even if the place is the same, may have to be used if sufficient accuracy is to be secured. So the average for a particular region may be used for that region in the rough rule. Let us now compute the constants using the data of Example 16 (a), and examine the degree of accuracy of the constants 5 and 10 used in the above verse. In the example we find that the time is 8 nāḍīs for shadow aṅg 13-56. The daytime for the day is nā. 33-20 and mid-day shadow, aṅg. 2-50, as we have already given in Example 17. Using the assumed form, x × 33 1/3 ÷ (13 14/15 + 2x − 2 5/6) = 8. x × 33 1/3 = 8 (2 x + 11 1/10) = 16 x + 88 4/5. 17 1/3 x = 88 4/5. x = 88 4/5 ÷ 17 1/3 = 444 × 3 ÷ (5 × 52) = 5 8/65. As 8/65 is small, x, the multiplier, may be taken as 5, and y (i.e. 2x) may be taken as 10, with tolerable accuracy, as VM himself seems to have done in the popular verse. Let us examine the accuracy given by this by working Example 17 using this. The time = 5 × 33 1/3 ÷ (13 14/15 + 10 − 2 5/6) = 500 × 10 ÷ (3 × 211) =nā. 7-54. Note how near this is to the correct 8 nāḍīs, and contrast with the result of the rule given by the text, nā. 8-37. Let us once again examine the relative accuracy by computing the time sought in the example under IV. 41-44, from the shadow caused by the Sun on the prime vertical, at the place and time of Example 16 (a). The prime vertical shadow was given as aṅg. 17-4. The time got there was nā. 6-53. Using the rule of the text, time = 6 × 33 1/3 ÷ (17 1/15 − 2 5/6 + 12) = nā. 7-37, which is far from the correct nā. 6-53. Using the popular verse, time = 5 × 33 1/3 ÷ (17 1/15 − 2 5/6 + 10) = nā. 6-53, agreeing exactly with the correct time. What are we to conclude from this? [नाडीतः छाया] छायाऽऽर्की नाडीभिर्दिनमानं षड्घ्नमुद्धरेत्तत्र । लब्धं द्वादशहीनं मध्याह्नच्छायया सहितम् ॥ ४९ ॥ Shadow from time 49. Roughly again, the shadow can be got thus from the time: Multiply the daytime by 6, and divide by the time for which the shadow is sought. Add the

IV. 50 IV. THREE PROBLEMS 123 mid-day shadow to the result and deduct 12. The shadow of the gnomon, caused by the Sun, is got. This means: Shadow at any time = 6 × daytime ÷ the time taken + midday shadow − 12. Example 19. Given daytime = nā. 33-20, mid-day shadow = aṅg. 2-50, find the shadow at nā. 8-0 from sunrise. Shadow = 6 × 33 1/3 ÷ 8 + 2 5/6 − 12 = 25 + 2 5/6 − 12 = aṅg. 15-50. (Actually the shadow is aṅg. 13-56, which can be seen from the previous examples). But if the constants in the popular verse, 5 and 10, are used, then the rule becomes, Shadow = 5 × daytime ÷ taken time + mid-day shadow − 10. Using this, the shadow = 5 × 33 1/3 ÷ 8 + 2 5/6 − 10 = 20 5/6 + 2 5/6 − 10 = aṅg. 13-40. See how close this is to the correct, 13-56. Being the inverse of the operation of finding the time from the shadow, this rule can be derived from the previous rule, viz, 6 × daytime ÷ (shadow − mid-day shadow + 12) = time in nāḍis. ∴ 6 x daytime ÷ time = shadow − mid-day shadow + 12. ∴ shadow = 6 x daytime ÷ time + mid-day shadow − 12, which is the present rule. [चन्द्रच्छाया] (इ)ष्टा नाड्यो द्युनिशं चन्द्रोदयनाडि[का]युतविही(नाः) । ताभिस्तत्कालेन्दोर्भानोरिव चिन्तये(च्छा)याम् ॥ ५० ॥ Moon's shadow 50. To compute the Moon's shadow at any time in the night, the time after sunset is to be added to the nāḍīs from moonrise to sunset if the Moon rises in the day. If the Moon rises after sunset, the time of moonrise after sunset is to be subtracted from the taken time. This is to be used as the time taken for computation, and work done as in the case of the Sun to get the Moon's shadow. The work is to be done thus: Upto the desired time after sunset, the time after moonrise is to be found, and this time is to take the place of the time after sunrise in the work of finding the shadow as in IV. 41-44. So, for the desired time the Moon's true declination and day-diameter have to be found and these are to be used in the place of the Sun's declination and day-diameter. The required cara etc. are to be found using these. As the nāḍis pertain to the solar day, they should be made lunar and used. The two examples given hereunder will make the work clear. The author's intention is 49. Quoted by Utpala on BS 2, p.62 49a. A. नाडिभि b. A. षट्समु. A1. द्वरेतत्र; A2. द्वारेतत्र 50a. A.C.D. दृष्टा. A. द्युनिशे b. A. नाडियुत. A1. विहीना; A2. विहिना c. A. कालेंदो d. A. छायां

124 PAÑCASIDDHĀNTIKĀ IV. 50 to convey that the inverse process of finding the time from the Moon’s shadow is also to be done as from the Sun’s. The time of moonrise required in this work will be given by the author in V. 8-10. The Moon’s true declination has been given already in IV. 16. The proof of the work is similar to that of the Sun’s. It must be remembered that in getting the time from the Moon’s shadow, succes- sive approximation has to be done, as in the case of the sun, for the same reason. The following should also be noted. If the desired time after sunset for which the shadow is sought is less than the time of moonrise after sunset, the work need not be done. Or if the moon sets in the night before the desired time, the work need not be done. Obviously, these should be examined before commencing the work. Much has to be said here, for which the reader is referred to works like the Siddhānta Śiromaṇi. Example 20. The sine of latitude of a place is 45' 56", and thence the sine of colatitude 110' 52". There, on a particular day the daytime is nā. 32-24. The moonrise is at nā. 27-18 after sunrise. At that time the Moon’s true declination is 15° south. (i.e. the Moon is in the southern hemisphere). Since Moon’s declination is 31' 4", and thence the day-diameter 231' 50". The cara-vināḍīs from these for the day is 132. The lunar day, i.e. the duration of moonrise to moonrise is 62 nāḍīs. Compute the shadow caused by the Moon at nā, 4-8 after sunset. The time to be taken for computation = the time from moonrise to the given time = the time from moonrise to sunset +the given time (after sunset) = nā. 32-24 – nā. 27-18 + 4-8 = nā. 9-14. The Moon’s cara-vināḍīs = 132, given. Both should be converted to the lunar measure. For 62 nāḍīs there is one lunar day, i.e. 60 lunar nāḍīs; so for nā. 9-14, there are 9-14 × 60/62 = 8-56 lunar nāḍīs. Converting into degrees, we have (8-56) × 6 = 53° 36'. Similarly, the cara-vināḍis made lunar = 132 × 60/62 = 128. Converted into degrees, 128/20 = 6° 24'. Now, using the rules of verses 40-44, (i) Sin altitude = {sin (53° 36' + 6° 24') – sin 6° 24')} × 231' 50" × 110' 52" ÷ 28,800 (the upper sign is taken because the Moon is in the southern hemisphere). = (sin 60° – sin 6° 24') × 231' 50" × 110' 52" ÷ 28,800 = (103' 55" – 13' 23") 231' 50" × 110' 52" ÷ 28,800 = 90' 32" × 231' 50" × 110' 52" ÷ 28,800 = 80' 49". (ii) gnomonic shadow caused by the Moon = 12√(14,400 – 80' 49"²) ÷ 80' 49" = aṅg. 13, vyaṅg 11. Example 21. For the same place and the same time of Example 20, find the time, given the shadow caused by the Moon is aṅg. 13-11, extending the method of verse 45-47 to the Moon. The required elements already given in Example 20 are: sin lat. 45' 56", sin colat. 110' 52", sin Moon’s declination 31' 4", sin Moon’s day-diameter 231' 50", time of moonrise nā. 27-18 after sun- rise, duration of the day nā. 32-24, and the duration of the lunar day = 62 nāḍīs.

IV. 51 IV. THREE PROBLEMS 125 (i) ‘First sine’ = 1,72,800 ÷ (110′ 52″ × √(13 11/60 + 12²)) = 87′ 26″. (ii) Earth-sine = 45′ 56″ × 31′ 4″ ÷ 110′ 52″ = 12′ 52″ (iii) Sine I = (87′ 26″ + 12′ 52″) × 240 ÷ 231′ 50″ = 103′ 55″, (since the Moon is in the southern hemisphere). (iv) Sine II = 12′ 52″ × 240′ ÷ 231′ 50″ = 13′ 23″. (v) Arc sine I = 60°. Arc sin II = 6° 24′. The time of shadow after moonrise = (60° − 6° 24′)/6 = 53° 36′/6 = nā. 8-56, (Moon being in the southern hemisphere). This time pertains to the lunar sāvana day, and converted into ordinary (i.e. solar) sāvana, the time after moonrise = 8-56 × 62 ÷ 60 = nā. 9-14. The time from sunrise = nā. 27-18 + nā. 9-14 = nā. 36-32. The time from sunset = nā. 36-32 − nā. 32-24 = nā. 4-8. The result is correct, because in Example 20, we took this same time and got the shadow aṅg. 13-11, which we have used in this example. चरनाडीक्रमविधिना द्युव्यासा (द्य) थामति [च] विक्षे (पात्) अस्तमयोऽप्यध्वविधिः शेषाणां युक्तितश्चिन्त्यम् ॥ ५१ ॥ 51. For the others, (i.e. for the luminaries other than the Sun and the Moon, viz. the star-planets) also, determining the corresponding operations, and using their respective latitude and day-diameter, and getting the cara-nādīs etc. (in terms of their respective sāvana days), (not only the work of finding the shadow for the given time and time for the given shadow as above, but also) their daily risings and settings and reduction to different localities should be thought out and done. The following is the idea. The computation of the rising and setting of the Sun has been given already in this chapter. The Moon’s rising and setting will be given below, in chapter V. Understanding the nature of the operation from these and taking the star-planets corrected to the different longi- tudes and computing their respective sāvana days and cara-vinādīs, using their latitudes to get their true declinations and day-diameters, everything done in connection with the Sun and the Moon should be done in connection with the star-planets also. It is from this that we understand that in the work of computing the Moon’s shadow we have to use the true declination, day-diameter, and time measured in the Moon’s sāvana day, as we have done already. Therefore this verse may also be taken as an extension of the previous verse. Here TS and NP have done a lot of emendations that are unnecessary for, without those emen- dations we get the same idea as they have given, at such pains. 51a. D. चरनाड्य [प] क्रमा [दि] विधिना A.C.D. om च. A.D. विक्षेपम्; C. विक्षेयम् b. A. द्युव्यासाम्यभाति; C. द्युव्यासाप्तक्रमादि; D. द्युव्यासाप्तक्रम c. C. मये पूर्व विधिः; D. मयेऽप्यूर्ध्वविधिः

126 PAÑCASIDDHĀNTIKĀ IV. 54 [छायातः दिक्साधनम्] छायाऽर्कवर्गयोगा (त्पदेन) भाज्यार्कसंगुणा त्रिज्या | विषुवज्जीवागुणिता (लम्बक) भक्ता तु सूर्याग्रा || ५२ || का (ष्ठ) घ्रयार्क्रमौर्व्या लम्बकहृतया वि (हीन) संयुक्ता | सूर्याग्रा (ऽज) तुलादौ कर्णघ्नी त्रिज्ययाऽपहृता || ५३ || लब्धाङ्गुलानि (को) टिस्ट (च्छा) यावर्गविवरमूलं [यत्] | स च (बाहुर्दिग्ग्र) हणे सममि (तिः) को (ट्या) तु देयमृजु || ५४ || Directions from shadow 52. Twelve times the radius (i.e. 1440) is to be divided by the 'Shadow- hypotenuse', i.e. the root of the sum of the squares of the shadow and 12. This multiplied by sine latitude and divided by sin co-latitude and divided by sin co-latitude is called Sūryāgrā (otherwise well-known as Śaṅkvagram or Śaṅ- kutalam). 53. From this Sūryāgrā, the sine of the Sun's declination divided by the sine of co-latitude (which is otherwise called Agrā) should be deducted or added, according as the Sun is in the six signs beginning with Aries, or the six signs beginning with Libra, (i.e. according as the declination is north or south). The result is to be multiplied by the 'Shadow-hypotenuse' and divided by the radius, (i.e. by 120). 54. What is obtained are termed Koṭi, (or 'Perpendicular'), measured in digits. The root of the square of the Koṭi deducted from the square of the shadow is called Bāhu (or 'Base'.) and the Koṭi is to be so constructed as to be perpendicular to the 'Base', (whose extension both ways is the prime vertical). Thus the directions are got. The following are the steps in the work: (i) Shadow hypotenuse = √(shadow² + 144) (ii) Sūryāgrā 12 × 120 × sin latitude ÷ (shadow hypotenuse × sin colat.) (iii) Agrā or Amplitude = sin max. dec. of Sun × sin Sun's long. ÷ sin colat. = sin dec. of Sun × 120 ÷ sin colat. (The declination is north if the Sun is in the six signs from Aries, and south otherwise) 52a-b. A.C.D. योगा (C.D. योगात्) पदे विभाज्यार्क      c. A. °ग्रा च तुलादौ   b. D. संगु(णिता] त्रिज्या |

IV. 54 IV. THREE PROBLEMS 127 (iv) (Sūryāgrā ∓ Agrā × shadow hyp. ÷ 120 = ‘Perpendicular’ (of ∓, the upper sign is for north declination, and the lower for south. If the ‘Perpendicular’ got is positive then it is north, if negative, south.) (v) √(shadow² − Perpendicular²) = Base Here, steps (ii), (iii) and (iv) can be simplified and put in the form: ‘Perpendicular’ = (12 × sine latitude ∓ shadow hypotenuse × sin declination) ÷ sin colat. (of ∓, the upper sign is for north decli- nation and the lower for south. As already said, the Perpendicular obtained is north if positive and south if negative. If, when the declination is north. Shadow hypotenuse × sin declination > 12 × sin latitude, then deduct the less from the greater and take it as negative, i.e. take the resulting Perpendicular as south.) C |
|
| \ Shadow | \ 5-0 Perp. 2-37 |
|
|
| \ Shadow angle A | 90° \ B +---------) 4-16 Base Fig. IV. 15 Example 22. The latitude of a place is 30°, whence sin lat = 60', and sin colat = 103' 55". The Sun at the time of taking the shadow = rāśi 1-15, whence sin Sun’s long = 84' 51", sin declination = 48' 48" × 84' 51", ÷ 120 = 34' 30", (north, as the Sun is in the first 6 signs). For this place and time if the shadow is 5 digits, find the direction of the shadow. (i) shadow hypotenuse = √(5² + 144) = 13. (ii) Sūryāgrā = 12 × 120 × 60 ÷ (13 × 103' 55") = 63' 57".2 (iii) Agrā = 48' 48" × 84' 51" ÷ 103' 55" = 34' 30" × 120 ÷ 103' 55" = 39' 51".4 (iv) ‘Perpendicular’ = (63' 57" − 39' 51") × 13 ÷ 120 = aṅg. 2-36.6 (The ‘Perpendicular’ is north, as the result is positive) (v) The ‘Base’ = √(5² − (2 − 36.6)²) = aṅg. 4-16. Or, using the simplified form, the Shadow-hypotenuse, 13 aṅg, being known, ‘Perpendicular’ = (12 × 60' − 13 × 34' 30") ÷ 103' 55" = 271' 30" ÷ 103' 55" = aṅg. 2-36.6. Then the ‘Base’ is calcu- lated as done above. Using the ‘Base’ and the ‘Perpendicular’, the direction of the shadow is found thus graphically. (see Fig. 15).

128 PAÑCASIDDHĀNTIKĀ IV. 54 Here AB is the ‘Base’ which, extended on both sides, is the prime vertical. AC is the ‘Perpendicular’, extending northwards from AB that lies east-west. Angle CAB = 90°. BC is the shadow, and angle ABC is the angle made by the shadow with the east-west line. The direction of the sun is the line CB extended backwards, making the same angle with AB extended. Example 23. For the same place and the same day, find the Sun’s direction, when the shadow is aṅg. 27-30. (i) Shadow hypotenuse = √(27½² + 144) = aṅg. 30 (ii) Sūryāgrā = 12 × 120 × 60 ÷ (30 × 103′ 55″) = 27′ 42″.8 (iii) Agrā = 39′ 51″.4, found in example 22. (iv) ‘Perpendicular’ = (27′ 42″.8 − 39′ 51″.4) × 30 ÷ 120 = aṅg. − 3-2, i.e. aṅg. 3-2 southward. (Or, which is the same, deducting 27′ 42″.8 from 39′ 51″.4, and doing the work with the remainder 12′ 8″.6, the perpendicular obtained is aṅg. 3-2, negative and ∴ southward). (v) ‘Base’ = √(27½² − 3 1/30²) = aṅg. 27-20. Or, by the simplified formula, ‘Perpendicular’ = (12 × 60′ − 30 × 34′ 30″) ÷ 103′ 55″ = (720′ − 1035′) ÷ 103′ 55″ = −315′ ÷ 103′ 55″ = aṅg. 3-2 southward. From the ‘Perpendicular’ the ‘Base’ is found as already done. The direction of the shadow is found graphically thus:

Base                                        B    To Sun
A +-----------------------------------------------+--------->
| 27-20                                        / Angle Shadow / Angle of Sun
P  |                                            /
e  |                                           /
r  | 3-2                                27-30 /
p. |                                         /
|                                        / Shadow
C +---------------------------------------+

Fig. IV. 16 Here, AB is the ‘Base’, which extended both ways, is the prime vertical. AC is the Perpendicular, directed southwards. BC is the shadow. Angle ABC is the direction of shadow. At an equal angle to the east-west on the opposite side is the Sun. Example 24. Sin lat of place = 72′ , whence sin colat = 96′. The longitude of the Sun = rāśi. 11-0, from which sin longitude of the Sun = 60′, and thence sin decl = 24′ 24″, south, since the Sun is within the six signs from Libra. Find the direction when the shadow is aṅg. 27-30. (i) Shadow-hypotenuse = √(144 + 27½²) = aṅg. 30. (ii) Sūryāgrā = 12 × 120 × 72′ ÷ (96 × 30) = 36′.

IV. 54 IV. THREE PROBLEMS 129 (iii) Agrā = 48′ 48″ × 60 ÷ 96′ = 30′ 30″. (iv) The Sun being in the six signs from Libra, ‘Perpendicular’ = (36′ + 30′ 30″) × 30 ÷ 120′ = aṅg. 16-37.5 north. (v) ‘Base’ = √(27½² − 16⅝²) = aṅg. 21-54. Or by the simplified formula, Perpendicular = (12 × 72′ + 30 × 24′ 24″)/96 = aṅg. 16-37.5. (+ is taken, as the declination is south). From this the ‘Base’ is calculated to be aṅg. 21-54 as before. The direction is graphically represented thus: [Figure IV. 17: Right-angled triangle ABC with right angle at A. Base AB labeled "Base" and "21-54", perpendicular AC labeled "Perp." and "16-37.5", hypotenuse BC labeled "Shadow" and "27-30". Angle B labeled "Shadow angle". Line CB extends backwards past B towards the southwest, labeled with an arrow "To Sun".] Fig. IV. 17 Here too, the angle of shadow is ABC, and the direction of the Sun is opposite to the shadow, making the same angle. We shall now prove the steps, taking them one by one: (i) Shadow-Hypotenuse: In the right angled triangle having the shadow as base and the twelve digit gnomon as perpendicular, the shadow-hypotenuse is the hypotenuse. Hence by the well-known formula, base² + perpendicular² = hypotenuse², √(shadow² + gnomon²) = shadow- hypotenuse. As the gnomon is 12 aṅgulas and the shadow too is measured in aṅgulas, the shadow- hypotenuse measured in aṅgulas = √(shadow² + 12²). (ii) Sūryāgrā: This is the distance between the line joining the rising and setting points and the diurnal circle (see Fig. 18). This is called śaṅkvagra by the earlier Bhāskara I and his followers and śaṅkutalam by the later Bhāskara II and others. It has been mentioned that, as seen from places on the earth other than the equator, since the circles on the stellar sphere are bent southwards (this is from the point of view of people in the northern hemisphere) the diurnal circles following these are also bent southwards. Therefore by the intersection of the arcs on the stellar sphere and the celestial sphere several right angled triangles are formed by their sine lengths, which triangles are called ‘Latitude-caused triangles’ (Akṣakṣetras). From the similarity of these triangles, when the length elements of one are known the corresponding length elements of another may be calculated by the rule of proportion. Among these, two similar triangles answer to our need, in one, which is well-known, sin lat is the base, sin colat is the perpendicular, and

130 PAÑCASIDDHĀNTIKĀ IV. 54 the radius is the hypotenuse; and in the other Sūryāgrā (i.e. śaṅkutalam) is the base, the Great gnomon is the perpendicular and what is called Taddhṛti is the hypotenuse (Vide Sid. Śiromaṇi, Gola, Tripraśna 49). Therefore, when sin lat, sin colat, and the Great gnomon are known Sūryāgrā can be calculated by the proportion: Sin colat : sin lat :: Great gnomon : Sūryāgrā. Sūryāgrā = Great gnomon × sin lat ÷ sin colat. The Great gnomon can be found from the similarity of the two triangles, in one of which the shadow is the base, the twelve-digit gnomon is the perpendicular and the shadow-hypotenuse is the hypotenuse, and in the other sin zenith distance is the base, the Great gnomon is the perpendicular, and the radius is the hypotenuse. Therefore by the proportion: shadow-hypotenuse : 12 :: radius : Great gnomon, the Great gno- mon = 12 × 120' ÷ shadow-hypotenuse. Hence by substituting we get, Sūryāgrā = 12 × 120' × sin lat ÷ (shadow-hypotenuse × sin colat). Since the celestial sphere is bent southward, Sūryāgrā is really south, permanently, (from the point of view of a man in the northern hemisphere, as we have already said). But here, as we are dealing not with the Sun but with the shadow, which is opposite to the Sun, we have taken the Sūryāgrā as always north. We shall illustrate these things by Fig. 18. We have mentioned that for observers in the northern hemisphere the diurnal circles bend southward, resulting in the 'southing' of the celestial bodies, because of the southward bend of the stellar sphere. As the shadow moves in the direction opposite to the Sun, the tip of the shadow moves in circles bent northwards, like I, II, III, in the Fig. Also, it should be remembered, as we are depicting the shadows in the Fig, the direction of Agrā and Sūryāgrā are reversed. Fig. IV. 18.

IV. 54 IV. THREE PROBLEMS 131 I: The circle on which the tip of the shadow moves on a day when the Sun is in the southern hemisphere. II: The circle on which the tip of the shadow moves on a day when the Sun is on the equator. III: The circle on which the tip of the shadow moves on a day when the Sun is in the northern hemisphere. A, B = rising and setting points of the Sun, on the day related to I. C, D = rising and setting points of the Sun on the day related to II, and E, F, related to III. AB, CD, and EF are the lines joining the respective rising and setting points and are parallel to one another. With reference to I, (i.e. for a day when the Sun is in the southern hemisphere), GA = HJ = Agrā (directed northward), JK = Sūryāgrā (directed northward) and HK = Agrā + Sūryāgrā, from which it is obvious that the Perpendicular is also directed northward. With reference to II, (i.e. for a day when the Sun is on the equator), the Sun rises at C itself and sets at D itself, and therefore the Agrā is zero. LM is the Sūryāgrā (directed northward) and the ‘Per- pendicular’ = Sūryāgrā ∓ Agrā, is also LM. With reference to III, (i.e. for a day when the Sun is in the northern hemisphere), Agrā = QP = NO (directed southward) and PR or OS is the Sūryāgrā (directed north). At a time sufficiently near sunrise or sunset, for which OS is the Sūryāgrā, the Perpendicular is NS (directed southward). This is the case where Agrā is deductive but numerically greater than the Sūryāgrā. At a time sufficiently near noon, for which PR is the Sūryāgrā, the Perpendicular is QR got by PR – PQ, QP being numerically less than PR. (iii) Agrā: This is the amplitude, and forms the distance between the parallel lines constituting the prime vertical and the line joining the rising and setting points. This is also the sine of the angles of the rising and setting points made from the East or West points, respectively. The author has given the formula for this in V. 39, without mentioning its name Agrā, as also here without men- tioning its name. The derivation of the formula has been given by us there. When the Sun is in the northern hemisphere, this is north, and when in the southern, it is south. But here, as we are dealing with the shadow, we have reversed the directions. One thing must be mentioned in this connection: TS and NP interpret the word Sūryāgrā as Agrā or ‘Sine of the amplitude of the Sun’, evidently assuming the derivation sūryasya agrā = Sūryāgrā, i.e. Agrā itself, because the context is the Sun here. As for Sūryāgrā itself, they simply call it ‘a sine’. They have failed to notice that if taken thus, the formula for getting them would become wrong. Even if somehow, by changing the order of words in the sentence, we make the formulae agree in this work, in the next work of getting the sun from the direction of the shadow, it would be impossible to secure agreement between the words there. But we must mention here that in the Mahābhāskarīya, Agrā is termed ‘Arkāgrā’, evidently by the derivation, arkasya agrā arkāgrā. Sūryāgrā is there called Śaṅkvagra, as we have already said. (Vide Mahābhāskarīya, III. 53-54). But here we have no choice except to go by the text. (iv) Perpendicular: From what we have already said, and from the Fig. 18, it can readily be seen that (Sūryāgrā ∓ Agrā) is the distance between the Prime vertical and the tip of the Great shadow. This is called ‘Bhuja’ by other authors. The Bhuja corresponding to the shadow is got from this by the proportion,

132 PAÑCASIDDHĀNTIKĀ IV. 56 Radius: 'Bhuja' :: shadow-hypotenuse : shadow-Bhuja, So we have, (Sūryāgrā ± Agrā) × shadow-hypotenuse ÷ 120 = shadow Bhuja. Our author calls this Koṭi or ‘Perpendicular’, as we have already said. But this does not matter, for in a right angled triangle, with the hypotenuse given (as here the shadow), the other two sides are perpendicular to each other, and any one may be taken as the base, and the other as the perpen- dicular. (v) Base: When the ‘Perpendicular’ is got from the well-known formula of the right angled triangle, Base² + Perpendicular ² = hypotenuse², (the shadow being the hypotenuse here,) we have, ‘Base’ = √shadow² − Perpendicular². Since the Perpendicular is north-south, the ‘Base’ is east-west, and is a part of the east-west line, as the foot of the shadow is on the east-west line. Since the east-west line is known, we can lay the ‘Base’ on it, lay the ‘Perpendicular’, at right angle, and draw the shadow. Clearly, if initially we have the shadow marked on the ground, we can get the directions by using this method. The angle between the shadow and the base gives the direction of the shadow. Obviously the direction of the Sun is given by the equal angle vertically opposite. What has been said here for the shadow may be said for the sun without reversing the direction as we have done for the sake of the shadow, and the Sun’s direction can be got. From that the direc- tion of the shadow may be got as being vertically opposite. But it is clear that the author says every- thing here for the shadow, and not for the Sun. [छायातः रव्यानयनम्] छायासमरेखान्तरगुणिता त्रिज्या स्वकर्णभक्ताऽस्याः । एकत्वे (विश्ले) ष्या सूर्याग्रा संयुताऽन्यत्वे ॥ ५५ ॥ लम्ब [क] गुणिता (भा) ज्या काष्ठामौर्व्या (ततो) ऽर्कः स्यात् । सूर्योद्गवेन विधिना ग्रहा (स्त) तोऽन्येऽपि कर्तव्याः ॥ ५६ ॥ Sun from Shadow 55. (Explanatory translation): By a process reverse to the previous one, the longitude of the Sun can be computed from the shadow, thus: Take the dis- tance of the tip of the shadow in aṅgulas, from the east-west line, multiply it by 120′, and divide by the aṅgulas of the shadow hypotenuse (mentioned in the previous work). This is ‘the sine’. (It may be seen that this is the Sūryāgrā ∓ Agrā, of the previous work). If the shadow is north of the east-west line then ‘the sine’ also is north. If the shadow is south, ‘the sine’ is south. Compute the Sūryāgrā as given already in the previous work. This is to be taken as always north (as already mentioned). If ‘the sine’ and Sūryāgrā are of different direc- tions, then ‘the sine’ plus Sūryāgrā is Agrā. (It must be remembered that they will be of different directions only when the Sun is in the northern hemis- phere, i.e. within the six signs from Aries). If they are of the same direction, then the Agrā is one deducted from the other. (If ‘the sine’ is greater, then the Sun is in the southern hemisphere, i.e. within the six signs from Libra. If

IV. 56 IV. THREE PROBLEMS 133 Sūryāgrā is greater, then the Sun is in the northern hemisphere, i.e. in the six signs from Aries). 56. The Agrā thus got multiplied by the sine of colatitude, and divided by 48′ 48″. is the sine of the Sun’s longitude and from that the sun is obtained. (From this sine, first the degrees of Bhuja, D, is got. If the Sun is in the northern hemisphere, then the Sun’s longitude is D, or (six signs − D). If the Sun is in southern hemisphere, the Sun’s longitude is six signs + D, or (twelve signs − D). What exactly it is of the diad must be determined by the Sun’s ayana). (Following the method for the Sun, the other grahas also can be got). The following are the steps in the work:– (i) As already seen, shadow-hypotenuse = √(shadow² + 144). (ii) As already seen, Sūryāgrā = 12 × 120 × sin lat ÷ (shadow-hypotenuse × sin colat). (iii) ‘The sine’ = the distance in aṅgulas from the east-west line to the tip of the shadow × 120′ ÷ shadow-hypotenuse. (If the shadow is north of the east-west line, ‘the sine’ is north, if the shadow is south of the east-west line, ‘the sine’ is south). (iv) (a) If ‘the sine’ is north, and greater than the Sūryāgrā, Agrā = ‘the sine’ − Sūryagrā, and the Sun is in the southern hemisphere. (b) If ‘the sine’ is north and less than the Sūryāgrā, Agrā = Sūryāgrā − ‘the sine’, and the Sun is in the northern hemisphere. (c) If ‘the sine’ is south, Agrā = Sūryāgrā + ‘the sine’, and the Sun is in the northern hemis- phere. (v) Sine longitude of the Sun = Agrā × sin colat ÷ 48′ 48″ = Agrā × sin colat × 5 ÷ 244. (vi) From the sine of longitude, the Bhuja degrees D, and using that the longitude of the Sun by examination, are to be obtained. As in the previous work, (iii), (iv) and (v) can be simplified thus: Sine sun’s longitude = (12 ×sin lat ± sin colat × the distance in aṅgulas between the tip of the shadow and the east-west line) × 150 ÷ (61 × shadow hypotenuse). In ± if the shadow is south of the east-west line then the upper sign is to be taken, and the Sun then is in the northern hemisphere. If the shadow is north, the lower sign is to be taken. In this case, if 12 × sin lat is greater, the Sun is in the northern hemisphere, and if sin colat × distance in aṅgulas, is greater, the sun is in the southern hemisphere. Example 25. Of a certain place, sin lat = 60′, sin colat = 103′ 55″. There, on a day during Uttarāyaṇā, when the length of the shadow is 5 aṅgulas, the distance of the shadow tip from the east-west line is measured to be aṅg. 1-36.6, north of the east-west line. Find the longitude of the Sun. (i) Shadow-hypotenuse = √(5² + 144) = 13 aṅg. 55a. A1. ॰न्वे तितेष्या; A2. ॰न्वे तिरतेष्या; A.D. सा ज्या D. ॰न्वेज्जारितैष्या b. A. काष्टा. A. मनोर्कः; D. हतार्कः d. A. सूर्याग्रा. A2. न्यवे c. A2. सूर्यो-वेन 56a. A. लम्बगुणिता d. A. ग्रहक्षतो

134 PAÑCASIDDHĀNTIKĀ IV. 56 (ii) Sūryāgrā = 12 × 120' × 60' ÷ (13 × 103' 55") = 63' 57".2 (iii) ‘The Sine’ = aṅg. 2-36.6 × 120 ÷ 13 aṅg = 24' 6". (This is north as shadow is north). (iv) As ‘the sine’ is north, the lower sign is to be used, i.e. the difference is to be found. There, as Sūryāgrā is greater, Agrā = 63' 57".2 – 24' 6" = 39' 51" (The Sun is in the northern hemisphere). (v) The sine of Sun’s longitude = 39' 51" × 103' 55" × 5 ÷ 244 = 84' 51". (vi) The Bhuja degrees D = Arc of 84' 51" = rāśi. 1-15. As the sun is in the northern hemisphere, the longitude is rāśi 1-15, or rāśi 6-0 — rāśi 1-15, i.e. rāśi 4-15. As the Sun is in Uttarāyaṇa, the longi- tude of the sun is rāśi 1-15. Using the simplified method, and taking the lower sign since the distance is north, sin Sun’s long = (12 aṅg × 60' ~ aṅg 2-36.6 × 103' 55") × 150 ÷ (61 × 13 aṅg.) = (720' – 271' 30") × 150 ÷ (61 × 13) = 84' 51". (As 12 × sin lat is greater, the sun is in the northern hemisphere). The rest of the work is the same. Example 26. Of a certain place, sin lat = 60', sin colat = 103' 55". On a Dakṣiṇāyana day, when the shadow is aṅg. 27-30, its tip is found to be aṅg 3-2.15 south of the east-west line. Find the Sun. (i) Shadow-hypotenuse = √(144 + 27½²) = aṅg. 30. (ii) Sūryāgrā = 12 × 120' × 60' ÷ (30 × 103' 55") = 27' 42".8. (iii) ‘The sine’ = aṅg. 3-2.15 × 120 ÷ aṅg. 30 = 12' 8".6 (south, as the shadow is south). (iv) As the sine is south, the upper sign is to be taken, and the Sun is in the northern hemisphere, and therefore, Agrā = 27' 42".8 + 12' 8".6 = 39' 51". (v) Sin longitude of Sun = 39' 51" × 103' 55" × 5 ÷ 244' = 84' 51". (vi) The degrees of Bhuja = Arc 84' 51" = rāśi 1-15 As the Sun is in the northern hemisphere, the longitude is rāśi 1-15 or rāśi 4-15. As it is Dakṣiṇāyana, the Sun is rāśi 4-15. Applying the simplified method for this, as the upper sign is to be taken, since the shadow tip lies south of the east-west line, sin long = (12 × 60 + 103' 55" × 3.2) × 150 ÷ (61 × 30) = 84' 51", and the Sun must be in the northern hemisphere. The rest of the work is the same as done already. Example 27. Of a certain place sin lat = 72', and sin colat = 96'. There, on a certain day in Uttarāyaṇa, when the shadow is aṅg. 27-30, the distance of its tip from the east-west line is aṅg. 16-37.5 north. Find the Sun. (i) Shadow hypotenuse = √(144 + 27½²) = 30. (ii) Sūryāgrā = 12 × 120' × 72' ÷ (30 × 96) = 36'. (iii) ‘The sine’ = aṅg. 16-37.5 × 120 ÷ aṅg. 30 = 66' 30", (north, as the distance is north).