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).

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

144 PAÑCASIDDHĀNTIKĀ V. 7
सविता यतः शशाङ्कात् कोट्या परिकल्पितस्ततः कोटिः ।
देयांशकाङ्गुलसमा भुजकर्णौ चाङ्गुलैरेव ॥ ६ ॥
शशिमध्यात् प्राक् कर्णः कोटिरतोऽतो भुजः शशाङ्कगतः ।
परिधावक्षो(न्ना)मः शौक्ल्यं मध्याद्धनुस्तत्र ॥ ७ ॥
6. The ‘Koṭi’ is to be drawn on that side of the Moon towards the Sun, north
or south, which is got in computing it, using the scale, one aṅgula = one
degree of ‘Koṭi’. The Bhuja and the Hypotenuse also should be drawn to the
same scale.
7. Thus, first there is the Hypotenuse from the centre of the Moon to that
of the Sun. From the centre of the Sun the ‘Koṭi’ is laid in the direction com-
puted for it. Then from its termination the ‘Bhuja’ is laid towards the Moon’s
centre. On the rim of the Moon represented by a circle of fifteen aṅgulas, the
raising of the horn in aṅgulas due to the latitude of the place is to be done. At
the centre of the two ends of the horn the illumination in digits is to be
represented on the diameter. There the arc (forming the upper boundary of
the illumination) is to be drawn (by making the arc pass through the two ends
of the horn and the point in the middle to which the illumination extends).
Though it is plain that these four verses give instructions for the graphical representation of the
Moon at the times of visibility, (specifically its first visibility in the evening in the west), yet on
account of possible incorrect copyings, and because we are not sure of the degree of roughness of
the result intended by the Siddhānta, we encounter a lot of difficulty in ordering the words and
interpretting them. The author has not given the diameter of the Moon in aṅgulas, which is neces-
sary to draw the orb, and represent in it the illumination and the uplifting of the horn. But we can
infer the diameter to be fifteen aṅgulas thus: On Aṣṭamī, at the middle of either fortnight, when the
hypotenuse is 90°, according to the rule for getting the illumination, we have 90/12 = 7½ aṅgulas
of illumination. We know that half the Moon is illuminated then, and therefore the whole Moon
should have a diameter of fifteen aṅgulas, as we have stated.
This agrees also with the ‘elevation of the horn’ due to the latitude, which can be shown thus: The
line joining the tips of the horn seen horizontal by a person on the equator, is seen vertical by a
person at the pole, i.e. at 90° latitude, because the celestial equator is inclined by 90° there, so as to
be coincident with the horizon. As the hypotenuse at the time for which the elevation is required
is small, we can take it that the elevation of the horā is proportionate to the degrees of latitude.
According to the rule for elevation given by the author, it is for 90° and 90 × 2 ÷ 15 = 12 aṅgulas,
along the rim of the quadrant, from the horizontal to the vertical. Therefore the whole rim, i.e. the
circumference, is 4 × 12 = aṅgulas and this shows that the diameter must be 48 × 7/22 = fifteen
aṅgulas very nearly. This agreement in the diameter, as calculated by the two rules, itself is a criterion
for the correctness of the rules.
7c. A.°वक्षोनामः; C.D.U.°वक्षो नाम
d. D.शौक्ल्यमध्यात्॰
6b. A.कोज्यापरिकल्पितकोटिः A.तदनु सूत्रं; D.तदनु [च] सूत्रम्’

V. 7 V. PAULISA MOON'S CUSPS 145
Now, we shall show why this elevation is always on the northern limb. As mentioned several times
before, when latitude is used in the rules given, it is always north latitude that the author means.
As seen from north latitudes, the circles on the stellar sphere are all bent towards the south above
the horizon. Therefore the hypotenuse also is inclined south, the angle of inclination being equal
to the latitude, the hypotenuse being small and taken as a straight line. By this inclination south,
the line joining the tips of the horns, which is perpendicular to the Hypotenuse is elevated in the
north and depressed in the south, the angle of elevation being equal to the latitude. This elevation,
measured on the rim in aṅgulas is, as we have shown, twice the latitude divided by fifteen.
In the matter of the addition or subtraction of the difference of declination and the Moon's
latitude, we have said that the author has in view only the visibility in the west in the evening, for
then alone is the statement correct. Perhaps the author thinks that this is enough, because the ele-
vation of the horn at evening appearance alone is observed anxiously by people, as an omen of
good or evil. Or the author thinks that the readers themselves will understand the reversal of addi-
tion and subtraction for the morning appearance, by analogy with what was done before in the case
of visibility. It must also be noted that the object here is only to represent the orb of the Moon as it
appears, and the Hypotenuse, Bhuja and Koṭi are given to serve this end. Therefore it would not
matter if these are represented on a different scale from that on which the Moon is given, as for
instance an aṅgula per degree here. (On this scale the Moon will have to be represented by a dia-
meter of a half-aṅgula.)
There is a view that the elevation of the horns should be observed when the orb of the Moon is
on the horizon. In that case, the Sun will be below the horizon, and the question of the difference
in scale will not arise at all.
So, these are the steps in the work:-
i. The elevation of the horn due to latitude in aṅgulas = latitude in degrees × 2 ÷ 15.
ii. Illumination or digit of illumination in aṅgulas = the difference in longitude in degrees ÷ 12.
iii. Koṭi in aṅgulas = diff. in declination in degrees ± latitude in degrees. (For evening in the west,
if the Moon's latitude and ayana are of like direction, addition, and if of different directions, sub-
traction. For morning in the east, reverse the addition and subtraction).
iv. Hypotenuse = aṅgulas equal in number to difference in longitude.
v. Bhuja in aṅgulas = √(Hypotenuse² – Koṭ i²).
vi. See fig. 5, below. On the surface on which the phenomenon is to be represented draw a hori-
zontal line and mark the north and south sides on both ends. Mark the point S on it to represent
the Sun. Mark a point A on the horizontal line on the side in which the Moon is situated, (this is
known when finding the Koṭi) such that SA = Koṭi. From A draw a perpendicular upwards equal
to the Bhuja and at the end mark M, the centre of the Moon. MS is the Hypotenuse. With M as
centre draw the orb of the Moon having a diameter of 15 aṅgulas. At M draw a diameter BC perpen-
dicular to the Hypotenuse. From the northern end the diameter, say C, measure the aṅgulas of ele-
vation due to the latitude of the place, along the rim, and mark the point D. Draw the diameter
DME. D and E are the tips of the horns. On the lower semicircle caused by DE, mark its mid-point,
F. Draw the radius FM. On this mark a point G, such that FG = the aṅgulas of illumination. Draw
the arc DGE by the well-known method of making a circle pass through 3 points. This is the upper
limit of the illumination. The figure of the Moon is now as it will be seen in the sky. The horizon is
between the Sun and the Moon, parallel to the original horizontal line. It must be remembered that