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 |
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| \ Shadow | \ 5-0 Perp. 2-37 |
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| \ 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).

IV. 56 IV. THREE PROBLEMS 135 (iv) As ‘the sine’ is north, the difference is to be taken. As ‘the sine’ is greater, Agrā = 66′ 30″ − 36′ = 30′ 30″, (and the Sun is in the southern hemisphere). (v) Sine latitude of Sun = 30′ 30″ × 96 × 5 ÷ 244 = 60′. (vi) The degrees of Bhuja = rāśi 1-0. As the Sun is in the southern hemisphere, the Sun is rāśi 6-0

• rāśi 1-0, i.e. rāśi 7, or rāśi 12-0 − rāśi 1-0, i.e. rāśi 11. As it is Uttarāyaṇa, the Sun’s longitude must be rā. 11. Applying the simplified method, since the lower sign is to taken as the distance is north, sin long = (12 × 72 ∼ 96 × 16-37.5) × 150 ÷ (61 × 30). Here since distance × sin colat is greater, Sin long = (96 × 16-37.5 − 12 × 72) × 150 ÷ (61 × 30) = 60′, and the Sun must be in the south- ern hemisphere. The rest of the work is the same. The proof of the above rules is as follows: In the previous work, the ‘Perpendicular’, i.e. the dis- tance of the tip of the shadow from the east-west line, was calculated, given the Sun and the shadow, and from that the ‘Base’ and the direction were calculated. Here, given the distance and the ‘Perpendicular’, the Sun is computed. Therefore this is the converse of the previous work, and can be derived from that. Steps (i) and (ii) are the same as steps (i) and (ii) of the previous work, and have been derived there. We shall therefore derive (iii), (iv) and (v) from (iii), (iv) and (v) there. In the previous work in (iv), ‘Perpendicular’ = (Sūryāgrā ∓ Agrā) × shadow hypotenuse ÷ 120. ∴ ‘The sine = (Sūryāgrā ∓ Agrā) = Perpendicular × 120 ÷ Shadow hypotenuse, as in (iii) here. Since, ‘the sine’ = (Sūryāgrā ∓ Agrā), when the Sun is in the northern and southern hemispheres, respectively, Agrā = Sūryāgrā ∼ ‘the sine’. It has been mentioned that Sūryāgrā is always north, ‘the sine’ is either south or north according to the line to the tip of the shadow from the east-west line, and Agrā is south if the Sun is in the northern hemisphere and vice versa. Therefore, when Agrā is north, (i.e. when the Sun in the southern hemisphere,) ‘sine’ is north, and greater than Sūryāgrā. Therefore, in using (‘the sine’ − Sūryāgrā), we get that the Sun is in the southern hemisphere. If Agrā is south, and therefore to be got negative by the addition of Sūryāgrā, (i.e. when the Sun is in the northern hemisphere), and ‘the sine’ is north, Sūryāgrā is greater than ‘the sine’. Here we have to use (Sūryāgrā − ‘the sine’), and we get that when the Sun is in the northern hemisphere. If Agrā is south again, (i.e. the Sun is in the northern hemisphere, again), and ‘the sine’ is also south, then we have the case, Agrā = Sūryāgrā + ‘the sine’, in which case also the Sun is in the northern hemisphere. From the Agrā, the sine of Sun’s longitude is got thus: In step (iii) of the previous work, Agrā = Maximum declination × sine longitude of the Sun ÷ sin colatitude. ∴ sin long. of the Sun = Agrā × sine colatitude ÷ max. dec. = Agrā sin colat ÷ 48′ 48″, as we get here in step (v). The explanation of getting the Sun’s longitude from its sine has already been given in connection with getting the sines for degrees (IV.1-15). Another point to be noted in this connection is this: In what the author gives, there is nothing to say about the addition of ‘the sine’ and Sūryāgrā when they are of different directions, and therefore

136 PAÑCASIDDHĀNTIKĀ IV. 56 about the Sun being in the northern hemisphere. But when they are of the same direction, and one is to be deducted from the other, the author mentions only the case where Sūryāgrā is to be deducted from the Sun (thereby assuming ‘the sine’ to be greater) the case in which the Sun is taken to be in the southern hemisphere. We have seen that when ‘the sine’ and Sūryagrā are of the same direction, the Sun is to be taken as in the northern hemisphere in the case (iv) in which Sūryagrā is greater, and ‘the sine’ is deducted from it. The author has omitted to mention this case. Has he forgotten it? We think not. He hopes that the reader himself will infer the changes to be made in this contingency, viz, that ‘the sine’ is to be deducted from Sūryagrā, and as the result is to be considered negative, and as the Agrā thus got is negative it is to be taken as south, and as south Agrā is for the Sun in the northern hemisphere, the Sun in the northern hemisphere will be inferred. As for TS and NP, here too they interpret Sūryāgrā as Agrā. They are not aware of the error that would be caused by this in the situation of the Sun, the hemispheres being reversed. For the matter of that they do not refer to the Sun’s situation at all, nor to the contingency of the reversal the sub- tractor and the subtrahend. [इति पञ्चसिद्धान्तिकायाम् वराहमिहिरविरचितायां करणाध्यायश्चतुर्थः]¹ Thus ends Chapter Four entitled ‘Three Problems: Time, Place and Direction in the Pañcasiddhāntikā composed by Varāhamihira

1. Col. A.C.D. इति करणाध्यायश्चतुर्थः

Chapter Five

PAULIŚA-SIDDHĀNTA — MOON’S CUSPS ५. पञ्चमोऽध्यायः पौलिशसिद्धान्तः — चन्द्रशृङ्गोन्नतिः Introductory In this chapter the Moon's visibility after or before heliacal setting, the appearance of its horns at the time of visibility with its geometrical representation, and the daily rising and setting of the Moon with its time of reaching the meridian are dealt with. We can surmise that this chapter is a part of the Pauliśa Siddhānta because the things required for the computations like the declination of the Sun and the Moon with the latitude of the Moon, are available to us only from the Pauliśa, the Romaka and the Saura having not been dealt with as yet, and because the methods are too rough to be attributed to the Saura. [चन्द्रदर्शनकालः] अपमान्तरसंयुक्तात् तदूनगुणिताच्छशाङ्करविविवरात् । मूलेनापमविवरे छिन्ने विक्षेपसंगुणिते ॥ १ ॥ फलमिन्द्वर्कविशेषाच्छोध्यं त्वयनानुकूलविक्षिप्ते । तद्व्यत्यासे देयं विपरीतं पूर्वसन्ध्यायाम् ॥ २ ॥ Time of Moon’s visibility

1. Find the difference in longitude of the Sun and the Moon, as also the difference of their declinations, (the mean declination of the Moon being used for this purpose). Multiply the sum of these two differences by their difference and find the square root of the product. By this ‘square root’ divide the product of the Moon’s latitude and the difference of declination already found.
2. The ‘result’ is to be subtracted from the difference in longitude, if visibility in the west is in question and the latitude and ayana (i.e. course northward or southward) of the Moon are of the same direction, or added to the difference in longitude if of opposite directions. If visibility in the east is in question, reverse the subtraction and addition. 1-3. Quoted by Utpala on BS 4.15. 1a. A.D. अयातान्तर \qquad\qquad\qquad\qquad\qquad\quad 2a. A. फलसिंध्वर्क \ \ b. A. ॰त्तद्वनयुक्ताछशांकविविरान् \qquad\qquad\qquad\quad b. A1. छोध्यचयनानु॰; A2. छोध्यं च यनानु॰ \ \ c. A.D. मूलेनायनविवरे \qquad\qquad\qquad\qquad\quad\ \ c. U. च्छेद्यमपमानुकूल

138 PAÑCASIDDHĀNTIKĀ V. 3 दिनकृत्सप्तमभवनात्तेनोदयनाडिकाद्वयं यदि वा | वियति विमले (तदे)न्दोर्लोकस्यालोक (आ)याति || ३ || 3. In the case of the visibility pertaining to the west, if a segment equal to the corrected difference in longitude takes at least two nāḍīs to rise in the east as reckoned by using the ascensional difference (for the place) of the seventh rāśi from the Sun, then the Moon will be visible, provided the sky is clear. In the case of visibility in the east, use the ascensional difference of the Sun’s rāśi itself. The following are the steps in the operation:- i. Find the difference in the longitude of the Sun and the Moon. ii. Find the difference of the Sun’s declination, and the Moon’s mean declination. iii. The square root = √[(diff. in long. + diff. in dec.) × (diff. in long. − diff. in dec.)] iv. Result = diff. in dec. × Moon’s lat ÷ the square root. v. Corrected diff. in long = diff. in long ∓ result. (Of ∓ the upper sign is to be used if visibility pertains to the west, and the latitude and the course of the Moon are of the same direction, or if the visibility pertains to the east and the latitude and course of the Moon are of different directions. The lower sign is to be used otherwise. vi. If the visibility pertains to the west, find the time of rising of an ecliptic segment equal to (v), by using the ascensional difference (for the place) of the seventh sign from the Sun and Moon. If it pertains to the east, use the ascensional difference of the Sun’s rāśi itself. If the time so found is greater than two nāḍīs, the Moon is visible; otherwise it is not visible. The time that we find in (vi) is the time of Moonset after sunset in the west, and the time of moon- rise before sunrise in the east. The sun, Moon, declinations and latitude of these times should be used and the work repeated for greater accuracy. Other siddhāntas mention this, though the author here has not done so specifically. Or, even before beginning the work, we can know the approximate times of moonset and moonrise, and do the work using the elements of these times. Near the time of new moon the Moon is invisible because the lighted up part is very small, and the sky itself is bright by the nearness of the Sun below the horizon. It has been fixed by the ancient authors by observation, that if the Moon sets within two nāḍīs after sunset, or if the sun rises within two nāḍīs after moonrise, the moon is not visible. (In places near the equator this criterion will be satisfied if the elongation of the Moon is in the neighbourhood of twelve degrees.) It is this we are finding by the computation, and it is obvious that the nearer the time of the elements used to the result, the better will be the result itself. Therefore is the need for successive approximation. If it is only for the sheer beauty of its appearance in the sky which has been described by poets like Kālidāsa, the first digit of the Moon is fit to be sought. But it is necessary for religious purposes 3b. A. ॰नोदया. c.A. तदिन्दो d. A.C.D. U. लोकमायाति ।

V. 3 V. PAULISA MOON'S CUSPS 139 also. The Baudhāyanas have to avoid Iṣṭi being performed on the day of the first appearance of the Moon, and do it on the previous day, and the offerings to the manes have to be done on the day pre- vious to the Iṣṭi. The Dharmaśāstras describe the seeing of the first digit of the Moon as meritorious. The Muslims consider their months ending with the first appearance of the Moon, and so this is important to them for calendrical purposes. The observance of the last digit of the Moon was neces- sary in ancient times, for from that they had to determine whether the same day or the next one would be the new moon day, so necessary for their religious rites. The importance can be guessed from the special names they had for the days at new moon, Sinīvālī and Kuhū in which the streak of the Moon will be visible and invisible, respectively. Example 1. At a certain place having lat. 30°N. examine the visibility of the Moon in the evening, given, the Sun at sunset = rāśi 1-0, the Moon at sunset = rāśi 1-15, and the Moon's latitude = 240' south. From the Sun and the Moon, their respective declinations are 704'N and 1004'N (mean). From the latitude 30°N, and Sun's declination the vināḍis of ascension at the place, of Scorpio, the seventh rāśi from Sun and Moon, can be calculated to be 355. From these, i. Diff. in long = rā. 1-15 − rā. 1-0 = rā. 0-15 = 15°. ii. Diff. in dec. = 1004' − 704' = 300' = 5° iii. The square root = √[(15° + 5°) × (15° − 5°)] = 14° 8'.4 iv. The Result = 5° × 4° ÷ 14° 8'.4 = 85'. v. Corrected diff. in long = 15° + 85' = 16° 25', (the lower sign, because the work pertains to the west (evening) and the Moon's latitude is south, while its ayana is north), vi. As the work pertains to the west, the seventh rāśi measure is to be used, which we have found to be 355 vināḍis. Using this, the time taken for 16° 25' to rise is 16° 25' × 355 ÷ 30° = 194 vināḍis. This is more than 2 nāḍis and so the Moon will be visible. As the time found is far above the requirement, we need not repeat the work using the elements of the time of moonset. Example 2. At a certain place (north of the equator) on a particular day in the evening the Sun rā 6-0. The Moon is rā. 6-15. The Moon's latitude is 4° 40'. The equinoctial shadow of the place is 4 digits. Examine for Moon's visibility. From the Sun, its declination is 0', and from the Moon its mean declination is 363' S. From the equinoctial shadow and the Sun's declination, the measure of the ascension of Aries, (seventh from Sun and Moon, since the computation pertains to the west) can be calculated to be 228 vināḍis. Using these, i. diff. in long. = rā 6-15 − rā 6-0 = 15° = 900' ii. diff. in dec. = 363' − 0' = 363'. iii. The square root = √[(900' + 363') × (900' − 363')] = 823'.5 iv. The result = 363' × 280' ÷ 823.5 = 123'.4 v. The corrected diff. in long. = 900' − 123'.4 = 12° 56'.6 (The upper sign because, the work pertains to the west, and the ayana and latitude of the Moon are of the same direction.) vi. As the work refers to the west, using the measure of Aries, the seventh rāśi from Sun and 12

140 PAÑCASIDDHĀNTIKĀ V. 3 Moon, the time for a segment equal to 12° 56′.6 to rise is, 228 × 12° 56′.6 ÷ 30 = 99 vināḍis. This is less than 2 nāḍis and so the Moon will not be visible that day. As the time got is far less than the requirement, repetition of the work is unnecessary. The steps are explained thus: [चित्र : Fig. V. 1 - Cel. Eq., D, M, M1 (M'), Ecliptic, Diurnal Circle, South, W, S, L, North] Fig. V. 1 Here, WD is the celestial equator. SM'C is the ecliptic and LM' is the diurnal circle of the Moon projected on the ecliptic. S is the Sun, M is the Moon and M' is the same projected on the ecliptic. MM' is the Moon’s latitude. SM' is the difference in longitude which is found in step (i). WS is the Sun’s declination, and DM' is the Moon’s mean declination. ∴ SL is the difference of the declinations, found in step (ii). Assuming the triangles as plane triangles, in the right angled triangle SLM', LM'² = SM'² − SL² = (SM' + SL) (SM' − SL) ∴ LM' = √(SM' + SL) (SM' − SL), and LM' being the square root, it is equal to √(diff. in long. + diff. in dec.) (diff. in long − diff. in dec.).... (step iii) M'C is the result and it is found thus: As MM' is perpendicular to SC, triangle MM'C is right angled at M'. ∴ angle SM'L = angle CMM' Therefore the two triangles are similar. ∴ M'C/MM' = SL/LM .

V. 3 V. PAULISA MOON'S CUSPS 141 ∴ M'C = SL × MM' ÷ LM', i.e. 'the result' = difference in declination × moon's latitude ÷ 'the square root', (which is step iv). Now for the additive or subtractive nature of 'the result': If the Moon's ayana is northward, i.e. if the ecliptic is inclined northwards (as in fig. 1), the Moon having south latitude, being at the end of a perpendicular to it, is lifted up. Therefore the Moon projected at M' is projected at C, as it were, and the difference in longitude which is the distance between S and M', is increased. So, in this case, 'the result' M'C is to be added. Now consider the case, when the ayana does not change, but the latitude also is north, like the ayana, as in Fig. 2. M¹ Fig. V. 2 C M South ← S L → North Now, clearly the Moon M at the end of M'M is bent downwards, with the result that M'C is deduc- tive in this case, as the instruction says. Let us next consider the case when the Moon's ayana is southward as in Fig. 3. C M M¹ Fig. V. 3 M C South ← L S → North Clearly in this case the Moon having north latitude is lifted up, and 'the result', CM', is additive, and the Moon having south latitude is depressed, and CM' is subtractive. Thus we have, for ayana and latitude having identical direction, 'the result' is subtractive and having different directions it is additive. This is for visibility in the west.

142 PAÑCASIDDHĀNTIKĀ V. 3 Now, for the visibility in the east: we are now looking eastward and successive points on the ecliptic are lower and lower towards the horizon. Therefore in figs. 1, 2 and 3, other things being the same, the ecliptic alone is to be represented as being directed downwards, as in Fig. 4. M¹ C M South S L North Fig. V. 4 Therefore, in each case taken up for consideration above, the direction of the ayana being changed, we see that for the ayana and latitude having different directions, 'the result' is subtrac- tive, and having the same direction it is additive. Thus step (v) is explained. Now for step (vi). We have already said that the Moon will be visible if it does not set within two nāḍīs after sunset, or if it rises before 2 nāḍīs before sunrise. (As visibility depends actually on other factors like the keenness of the eyesight of the observer, we have only to take the authority of the Śāstras in this matter). So in the evening we have to find the time by which the Moon will set after sunset, i.e. the segment constituting the corrected difference in longitude will set. As the distance between the rising and setting points in always in 6 rāśis, this time is equal to that of the rising of an equal segment in the east, which can be calculated by using the vināḍīs of the ascensional difference of the rising sign, which being six rāśis away, is the seventh from the Sun (or Moon). If this time is greater than 2 nāḍīs, the Moon would not have set, and therefore be visible. In the matter of visibility in the east, the same explanation holds, except that now the time of rising of the segment in the east is wanted, using the ascensional difference of the rising sign in which the Sun (or Moon) itself is situated, and hence the instruction to use that sign. This instruction to use the ascensional difference of the same sign as the Sun in the case of visibility in the east is implied by the use of the word vā, though not explicitly stated, and can also be inferred from the nature of the explanation. TS-NP do not seem to have noted the difference in the methods to be pursued in the operation. Another mistake they have made is that they have discarded the correct reading, ayanānukūlavikṣipte (v. 2) and chosen the incorrect reading apamānukūlavikṣipte and accordingly, have given the condition for additiveness or subtractiveness, "If the moon's latitude is

V. 5 V.-PAULISA MOON'S CUSPS 143 of the same direction as the difference in declination etc." Declination had direction, but what direction can be attributed to the difference in declination as given in the text? Or how can the word for declination mean difference in declination? Whatever the latitude, 'the result' is zero at the junction of the ayanas, which means its sign, i.e. its additiveness or subtractiveness changes there, and therefore the ayana should be a criterion for additiveness or subtractiveness. The very name of this correction, Āyanadṛkkarma (this name is not mentioned here by the author, but it is this) will suggest that the ayana of the Moon must form part of the criterion. Another thing must be mentioned: The work given here is very rough, because spherical tri- angles are taken as plane triangles, and another correction called Ākṣadṛkkarma which is to be done for the sake of the latitude of the observer has been omitted. Therefore the reader should refer to works like the Mahābhāskarīya and Siddhānta Śiromaṇi for greater accuracy. [चन्द्रशृङ्गोन्नतिः तत्परिलेखाश्च] द्विगुणेऽ(क्षे) 'तिथ्यंशः' शृङ्गमुदक् तुङ्गमुडुगणाऽधिपतेः । देयं च भुजादेतच्छौक्ल्यं कर्णाद् द्विषट्कांशम् ॥ ४ ॥ अपमान्तरविक्षेपा(वे)कान्यत्वे युतोनितौ कोटिः । कर्णो रवीन्दुविवरं तत्कृतिविवरात् पदं बाहुः ॥ ५ ॥ Diagram of the Moon's cusps 4. Multiply the latitude of the place in degrees by two and divide by fifteen. By the resulting number of aṅgulas or digits (measured along the rim), the northern tip of the horn of the Moon should be raised upwards (as caused by the latitude at the time of first visibility). This raising should be directed upwards like the 'Bhuja' which we are going to mention. The number of digits of illumination of the Moon's orb, (usually called merely digits), is the twelfth part of the difference in longitude in degrees, last found, and should be directed like 'Hypotenuse', which we are going to mention. 5. The difference in declination last found should be added to the Moon's latitude or subtracted from it, as the directions of the Moon's ayana and its latitude be the same or different. (This refers to the visibility in the west in the evening. With reference to the visibility in the east in the morning, the addi- tion and subtraction, is done vice versa). The result is called 'Koṭi'. The differ- ence in longitude is called 'the Hypotenuse'. The 'Bhuja' is the square root of the difference of the squares of the 'Hypotenuse' and the 'Koṭi'. 4-7. Quoted by Utpala on BS 4.15. 5a. A.अनान्तर; C.D.अयनान्तर. A1.विक्षेपा; 4a. A.द्विगुणेच्छे; C.U.दिनगुणेच्छा; D.द्विगुणाक्षे A2.धिक्षेपा b. A.शृंगमुदकुंमुदुगुणाधिपतिः b. A1.2.वैकानले; A2.वैकानचे U.वैकान्यत्वे d. A.कर्णाद्विष्टकांश: D.कर्णाद्विष्टकांशः A.यातोनिता; C.युतोनिता c. A.रवींदुविवरं