सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

302
Comm. Ref. fig. 52. I.A. = AN. I.A. - Carajyā =
B.N. Hvers⁻¹ (BN) = arc EN. Arc EN + Caracāpa =
EN + EM = MN. This converted into time is evidently
the Unnatakāla because the arc MN of the equator is the
arc intercepted between the feet of the declination
circles at rising and at the time concerned M being the
foot of the rising declination circle and N the foot of that
at the time in question. The convention of signs is
clear.
Verse 70 and first half of 71. To obtain the Sun's
longitude from the shadow S.
The gnomonic shadow at noon, being multiplied by R
and divided by K, the inverse H sine of the result gives
the meridian zenith-distance. This being decreased or
increased by the latitude gives the Sun's declination
according as the extremity of the shadow is north or
south. From the declination, we have the Sun's longitude
by the formula H sin δ = (H sin λ H sin ω) / R .
Comm. We have from the triangle formed by the
gnomon and the shadow S, S / K = sin z or SR / K = H sin z
∴ H sin⁻¹ (SR / K) = z. Since we are directed to take
the mid-day shadow, we have the meridian zenith-distance
and from the formula z + δ = φ we have δ. If the
extremity of the shadow be north, the Sun is south of
the zenith, and then φ ~ z = δ. The word वियुक्ताः is
used to signify difference which is positive. If z > φ
then the declination is south and vice-versa. If the
extremity of the shadow is on the south, the Sun is on
the north of the zenith. in which case φ + z = δ.
Second half of verse 71. To obtain φ from δ.

308
If the zenith-distance and the declination are of the
same direction, their difference, otherwise their sum will
be the latitude.
[Fig. 58] Comm. Suppose the zenith-
distance is north and decli-
nation also north, then
clearly from fig. 58, in this
position S₁ of the Sun QS₁ —
ZS₁ = ϕ = δ — Z (1) Again
in the position S₃ of the Sun,
zenith-distance is south and
declination is also south ; so,
here also difference gives ϕ
ie. ZS₃ — QS₃ = Z — δ = ϕ.
(3) In the position S₂, how-
ever, Z is south and δ is north, so that their sum is
equal to ϕ. It will be noted that the Hindu convention
of signs does not contemplate negative declination and
also it uses the word 'difference' to signify the positive
difference, as for example, in the first two cases δ ~ Z is
taken as ϕ. The modern formula Z + δ = ϕ applies
universally with the convention that δ is +ve if north,
ϕ is +ve if south.
Verses 72, 73. To obtain the Bhuja from the shadow.
Karṇa Vrittāgrā = (A × K) / R where A = Agrā and K
the Chāyākarṇa. This Karnāgrā is to be taken as
belonging to the opposite hemisphere to the Sun. Calling
this Karnāgra as a and the equinoctial shadow as s, a ⨦ s
according as δ is south or north gives the Chāyābhuja b.
Thus a ⨦ s = b. If the extremity of the shadow be on the
north and δ be north, then b + a = s ; if δ be south
b ~ a = s. If b be north, b ~ s = A, otherwise ie. if
south b + s = a. (R × a) / K = Agrā and
(Agrā × Koṭi of latitudinal triangle) / (Karṇa of the latitudinal triangle) = H sin δ.

304
Comm. These verses are very important and the
contents have been already elucidated under verses 13–17.
We shall elucidate the convention of signs in more detail
both from the modern point as well as from the Hindu
traditional point. First we shall discuss the modern. We
have the formula A = S + B where A = Agrā, S =
Sankutala and B, Sanku-bhuja. Let us confine ourselves
to north latitudes alone, for, south latitudes did not
concern the Hindu astronomers. Then treating north
declination as positive and also north Hindu azimuth as
positive the above formula holds universally. (Ref. fig. 59)
Let the figure represent the meridian plane. Let S₁, S₂, S₃
be the projections of the Sun's positions in their diurnal
circles on to the meridian plane. Let A, B. be the
Fig, 59

305
projections of the rising points of the Sun on the same
plane. Let perpendiculars be dropped from S₁, S₂, S₃ on
the plane of the horizon. Let o be the centre of the
sphere. Let ns be the north-south line. In position S₁,
S₁C = Śanku-Bhujā, KA = Śankutala, oA = Agrā, so
that A = S + B (1) In the position S₂. S₂D = Śanku-Bhuja,
LA = Śanku-tala, oA = Agrā, so that B + A = S; but
here B is negative, a the azimuth being south so that writ-
ing - B for B, - B + A = S ∴ A = S + B again. In
position S₃, OB = A, MB = S, OM = B so that A + S = B;
but here A is +ve, δ being south and B is negative a
being south; hence writing - A and - B for A and B
A + S = - B or A = S + B again. This shows that
with the convention cited above A = S + B holds good
universally.
Now let us consider the situation with respect to
the Karnāgrā. Each of the three quantities a, b, s have
now opposite directions. If δ be north, the Sun will be on
the north of Equator, but the extremity of the shadow will
now be on the south of the Equinoctial shadow line
(E.S.L.) ie. the line which is parallel to the East-west
line at a distance of the equinoctial shadow s, and which
is tbe locus of the extremity of the shadow on the
equinoctial day; thus when the Agrā is considered to be
positive being on the north of the East point, the
Karnāgrā, though it is on fhe south will have to be
considered positive. Similarly when the azimuth of the
Sun a is on the north of the prime-vertical and is so
considered to be positive, the extremity of the shadow
being on the south of the East-west line and has a
negative azimuth, the bhuja is still to be considered
positive. Again the Śankutala being always south of
the Udayāsta Sūtra being considered south and positive,
the corresponding quantity into which it gets converted
on the horizontal dial namely the equinoctial shadow s
will be north of the East-west line and will be considered
39

306
positive. In other words in the equation a = b + s, a
is positive when δ is north, b is +ve when the azimuth of
the Sun is north of the East point, and s is always
positive. Since in north latitudes, Śaṅkutala will be
always south of the Udayāstasūtra and considered positive,
the E.S.L. will be on the north of the East-west line so
that s is considered positive. We should have had to
consider s negative in southern latitudes, as per the above
convention but as the Hindu astronomers did not have
to concern themselves with south latitudes, the question
of sign for s did not arise except taking it as always posi-
tive. Hence the equation a = b + s will hold universally
with the same conventions of sign which we stipulated
with respect to the equation A = B + S. The foregoing
analysis is on the modern lines.
[चित्र: वृत्त के अंतर्गत W-E व्यास, E·S·L समानांतर रेखा, तथा बिन्दु O, C, P, R, M, L, A, N, B, K युक्त ज्यामितीय आरेख]
Fig. 60

307
Now let us see how the convention of signs is
stipulated in Hindu astronomy with respect to the equation
a = b + s. We have said that ‘s’ is always north of
the East-west line and considered positive. Regarding
‘a’, it is said by Bhāskara व्यस्तगोला which means
that when δ is north and the Sun is said to be in
northern hemisphere, a is said to belong to the southern
hemisphere. Also when the Sun is on the north of the
prime-vertical and Śankubhuja is considered north, the
Chāyābhuja being south of the East-west line is considered
south. With these conventions of directions (we say of
directions, and not signs because the Hindu astronomers
do not speak of signs but only of directions) it is stipulated
in Hindu astronomy that quantities of like directions are
to be added, otherwise their difference is to be taken
‘तुल्यदिशोर्योगः, भिन्नदिशोः अन्तरम्’. This convention stipul-
ating addition or difference is technically called ‘संस्कार,
Samskāra’. That is why it is said simply “पलच्छायया
सौम्यया संस्कृता” ie. ‘Samskāra (on the aforesaid lines)
is to be effected between a and s to get bhuja b’. Here it
may be reiterated that the word ‘अन्तरम्’ ie. ‘difference’
is used in its restricted sense namely that the positive
difference alone is to be taken. Thus the ‘antaram’ of
8 and 5 is 3 as well as that of 5 and 8 is also 3. Then it
might be asked how to decide the direction of the bhuja, if
we were to take s as equal to a ~ b and not a-b or b-a.
The answer is that between a and s whose directions are
known as per the aforesaid convention, equating their
difference ie. a ~ s to b, we have to take b as having that
direction which is indicated by that quantity either a or s
which has a larger numerical value.
Thus while on modern lines we take a = b + s to hold
universally with the convention of signs which we have
agreed to on modern lines namely that a is +ve if δ is
north and b is positive if the Hindu azimuth is north of
the East point and s is always +ve, we take on the Hindu
lines a ± s = b with the conventions stipulated with respect

308
to directions (not of signs) namely that a is south if δ is
north, s is always north and b is to be taken to belong to
that direction to which the numerically bigger quantity of
a and s belongs in the case of difference. Also it is to be
taken to belong to that direction of a and s when both of
them have the same direction.
With this convention in mind, Bhāskara clarifies the
convention by citing examples.
(1) S=5, δ is north, Agrā=916'-48'' K=30 so that
a = KA / R = (916⅘ × 30) / 3438 = 8 units (south, because it should
be taken to be व्यस्तगोला ie. belonging to the direction
opposite to that of δ).
Question. "What is b and of what direction?"
Answer. b = a +~ s = 8 (south) ~ 5 (north). (We are
taking the difference because Samskāra is to be construed
as addition of quantities having the same direction and
difference of quantities of opposite direction ∴ b = 3
and is on the south because the numerically bigger quantity
of 8 and 5 belongs to south.
Q. 2. δ is north S=5, A=916'-48'' ; K=15, 'what
is b and in what direction?'
Answer. a = KA / R = 4; b = a +~ s = 4 +~ 5 = 4~5 (here
a is south δ being north and s is north so that difference
is stipulated as above) = 1 (north because 5 belongs to
north.
We add here two more examples to illustrate addition
by saying that δ is south in the above examples so that a
is north in both the examples. Hence in (1) b=8+5=13
(north) and in (2) b=4+5=9 (north). Refering to Fig. 60,
we see there three cases depicted namely the extremi-

309
ties of the shadows being A, B and C. In the first case
AL = Karnāgra = a (Karnāgra is the distance of the
extremity of the shadow from the E.S.L. namely K.L.R.
whereas bhuja is the distance of the same from the East-
west line namely PMN) AM = bhuja and ML = s so that
b=a+s addition being justified since both a and s are of
the same direction namely north. In the second case
BK = Karnāgra, BN = bhuja and NK = s so that b=s-a,
the difference being justified because a is south and s north.
Here we have taken the difference as s-a and not a-s
because s is numerically greater and being oriented north,
the bhuja is north. In the third case, PR=s, CP=b and
CR=a so that b=a-s, the difference being justified be-
cause s is north and a south. Also we have taken the
difference as a-s and not s-a because a is numerically
greater and as such lends its direction namely 'south' to
the bhuja. Thus in the three examples cited, addition is
prescribed between a and s only when the extremity of
the shadow is to the north of E.S.L. In the case of
A=S+B or B=A-S also, addition is prescribed only when
δ is south, which means that the corresponding a ie. AL
is north. In fact the prescription of addition or differ-
ence accord in the cases of both the equations either
A=B+S or a=b+s.
Verses 74 and 75. Hereafter questions are being set
and answered on diurnal problems.
Seeing the shadow of the gnomon, the azimuth and
longitude of the Sun or seeing two shadows with their
respective directions, whoever knows the equinoctial
shadow of the place, I consider him as the Garuda or Eagle
who could overcome the false pride of puffed up snakes of
astronomers.
Given that when K=30 units, the bhuja is 3 units
south, and when K=15, the bhuja is 1 unit north, com-
pute the latitude, or again given H sin δ = 846 and given
K and b of a shadow, compute the equinoctial shadows.

310
Comm. The questions are clear the second verse
illustrating the first.
Verse 76. Answer of the first question.
(b₁ K₂ ⁺~ b₂ K₁) / (K₂ ~ K₁) = s according as the bhujas are of the same
or opposite directions.
Comm. Suppose b₁ and b₂ are of the same direction so
that b₁ = a₁ - s and b₂ = a₂ - s, taking the modern convention
of signs. But a₁ = (K₁A) / R and a₂ = (K₂A) / R
∴ b₁ = (K₁A) / R - s and b₂ = (K₂A) / R - s
∴ (b₁ + s) / K₁ = A / R = (b₂ + s) / K₂
∴ K₂ b₁ - K₁ b₂ = s (K₁ - K₂)
∴ s = (K₂ b₁ - K₂ b₂) / (K₁ - K₂)
If, however b₁ and b₂ are of opposite directions ie. of
opposite signs, writing - b₂ for b₂, we have
s = (K₂ b₁ + K₁ b₂) / (K₁ - K₂) . Here we have chosen to follow the
modern convention of signs; otherwise we have to con-
sider four alternatives, for, bhujas of the same direction
might mean both of the type OC (fig. 60) or both of the
type of OB or one of the tyye OB and one of the type of
OA or both of the type OA.
Verses 77 and 78. Answer to the second question.
Let Laghu ≡ L = ((KH sin δ) / R)² ; 12² (L - b²) ≡ Ādya ;
Para = 12² b. Let Ādya and Para be divided by L ~ 12² ;
call them still Ādya and Para; then √(Para² + A) ± Para
= s according as b is north or south.

311
Comm. The data are
H sin δ, K and b; since K
and b are given a is known.
Thus from the triangle PZS
(fig. 61) we have sin δ =
sin ϕ cos z + cos ϕ sin z sin a,
all quantities except ϕ
are known. Solving this
trigonometrical equation
which is of the form a cos ϕ

• b sin ϕ = c, we can
  have ϕ.
  Fig. 61
  We shall now see how it is solved by Bhāskara. Let
  s be the equinoctial shadow. Then
  a = b + s = KA / R = (KH sin δ) / (H cos ϕ) I
  But 12 / K = 12 / √(s² + 12²) = (H cos ϕ) / R so that
  H cos ϕ = 12 R / √(s² + 12²). Substituting in I
  b + s = (KH sin δ √(12² + s²)) / 12 R
  which reduces to 12² R² (b + s)² = K² H sin² δ (12² + s²)
  ie. s² (12² R² − K² H sin² δ) + 2b 12² R² s =
  12² {(H sin² δ) K² − b² R²}
  ie. s² (12² − (K² H sin² δ) / R²) + 2. 12² b. s) = 12²
  ((K² H sin² δ) / R² − b²)
  Here (K² H sin² δ) / R² is symbolized as L
  12² ((K² H sin² δ) / R² − b²) put as Ādya and 12² b is put as para.

312
So the equation reduces to
s² (12² - L) + 2. Para. s = Ādya ; Divide throughout by
12² - L and put again Para / (12² - L) as Para and Ādya / (12² - L) =
Ādya Then the equation reduces to s² + 2 Para s =
Ādya ; completing the square (s + Para)² = Para² + Ādya
∵ s + Para = √(Para² + Ādya)
∴ s = √(Para² + Ādya) - Para as one solution.
We have taken to start with a = b + s which holds
good according to the Hindu convention when b is
south ; if, however b is north a = b ~ s so that (b ~ s)² =
s² + b² - 2bs. So in the equation we have to write -s
for s, so that we have now s² - 2 Para s = Ādya ie.
(s - Para)² = Para² + Ādya
∵ s = √(Para² + Ādya) + Para as the second solution.
Verse 79. When the Sun's longitude is 135°, the
shadow of the gnomon is 12 units and west. What is the
latitude?
Comm. Here is a method of obtaining the latitude
of the place by observing the gnomon's shadow when the
Sun is on the prime-vertical.
Verse 80. Answer to the question above.
(12 R) / K = H cos z ;
s = (12 H sin δ) / √(Sama-Sanku² - H sin² δ)
Comm. Solution in modern terms.
S = 12 ∴ tan z = 1 ∴ z = 45; but when the
Sun is on the prime-vertical, we have by Napier's rule
Sin δ = sin ϕ cos z = sin ϕ √2. But since λ = 135°

often handles vertical fractions in plain text either as:
[L..] triangle (H sin δ) / x = s / 12 where x is the Koṭi
OR
line-by-line exact layout.
Wait! Let's examine how the lines flow:
If we look at:
triangle H sin δ / x = s / 12 where x is the Koṭi
Let's check the rest of the page:

• ∴ s = (12 H sin δ) / x. But we are given that the Sama-
• Śaṅku ie.
• H cos z = R cos z = R / √2 because z = 45° when the
• shadow equals the length of the gnomon.
• and H sin δ = (H sin λ H sin ω) / R = (H sin 135 H sin ω) / R
• = (H sin 45 H sin ω) / R = R/√2 × (H sin ω) / R = (H sin ω) / √2
• ∴ x = √(Sama-Śaṅku² - H sin² δ) = √( R²/2 - (H sin² ω)/2 )
• `= (H cos ω) / √2 ∴ s = (12 H sin δ × √2) / (H cos

314
that moments I would reckon you as one who could be
well compared with the goad that could be applied to the
head of the wild elephants of puffed up astronomers.
Verses 82, 83. Answer to the first question.
Assume the H sine of the Unnatakāla to be Iṣṭa Hṛti
in the first place. Multiply it by 12 s and divide by k² the
square of the Viṣuvatkarṇa. Then you get an approximate
value of H sin δ. Then compute with this, H sin δ,
Charajyā etc. and thereby obtain a more correct value of
the Iṣṭa Hṛti. Multiply this by the H sin δ got before
and divide by the first Hṛti assumed. Then we have a
nearer approximation of H sin δ. Repeat the process till
a stationary value has been reached. That will be the
correct H sin δ.
Comm. We know that H sine of the Unnatakāla is
nearly the Iṣṭāntyakā. It will be noted that Iṣṭāntyakā
is the sum of two H sines namely (1) Carajyā (2)
H sine of Unnatakāla minus Chara. The second H sine is
called Sūtra or H cos h. (Vide page 278).
Thus Sūtra + Carajya = Iṣṭāntyaka whereas
H sine (Unnatakāla) = H sine of the Cāpas of Carajyā and
Sūtra. In other words H sin (Unnatakāla) =
H sin (H sin⁻¹ Carajyā + H sin⁻¹ (Sūtra)). Iṣṭa Hṛti is
H cos δ
Iṣṭāntyaka × ----------- . But we do not know H sin δ so
R
that Iṣṭa Hṛti could not be got. So we will not be far
from truth in assuming the given Unnatakāla to be Iṣṭa
Hṛti itself ie. Taddhṛti here, as the Sun is on the prime-
vertical. The formula for Taddhṛti is
R² H sin δ R². H sin δ K² H sin δ
----------------- = --------------------- = ------------
H cos ϕ H sin ϕ R. 12 s 12 s
----- × R × ---
k k

315
∴ (Taddhṛti × 12 s) / k² = sin δ. Thus assuming H sine
of the given Unnatakāla to be Taddhṛti and multiplying
it. by 12 s and dividing by k² we have the value of H sin δ.
But this is approximate because the given Unnatakāla is
not exactly Taddhṛti but only an approximate value.
From this H sin δ, compute H cos δ, Charajyā, Kujyā
and through the process indicated in verse 54 namely
“Subtract the Characāpa from the Unnatakāla. The
H sine of the result is called Sūtra. Multiply the Sūtra
by H cos δ and divide by R; then we have Kalā. Add
Kujyā to Kalā; we get Iṣṭa-Hṛti”, we obtain a more
correct value of Taddhṛti. Then here we may cut short
the process as follows namely ‘ If by the assumed Taddhṛti
we had the previous H sin δ, what shall we have for this
more approximate Taddhṛti ’? The result will be a more
approximate value of H sin δ. Again form the Taddhṛti
with this H sin δ and so repeating the process till we have
a stationary value, we have the correct value of H sin δ.
Note. This is a beautiful example of the method of
successive approximations which is a modern technique
but which was so much in vogue and favourite with the
Hindu astronomers. (It will be noted how to cut short the
method).
Verses 84, 85. Answer to the second question.
Obtain 12² R²/(R² − H sin² h) s² + 1 and divide R² by
this and take the square root which gives H sin δ. Then
(R × H sin δ) / (H sin ω) gives H sin λ whose Cāpa gives the longitude
of the Sun.
Comm. The H sine of the given Natakāla is H sin h
and R² − H sin² h = H cos² h. Let H sin δ be x, which
is required to be found. Then R² − x² = H cos² δ;
H cos h = Sūtra and

316
(Sūtra × H cos δ) / R = Kalā = (H cos h . H cos δ) / R = ∴
(H cos h . √(R² — x²)) / R
But Kalā is the Koti of the fifth latitudinal triangle of which
H sin δ is Bhuja. Hence (Kalā × H sin ϕ) / (H cos ϕ) = H sin δ = x
ie. (H cos h √(R² — x²)) / R × (H sin ϕ) / (H cos ϕ) = x; but (H sin ϕ) / (H cos ϕ) = s / 12
∴ Squaring both sides [(R² — H sin² h) (R² — x²)] / R² × s² / 12² = x²
∴ 12² R² x² = s² R² (R² — H sin² h) — s² x² (R² — H sin² h)
ie. x² {(12² R² + s² (R² — H sin² h)} = s² R² (R² — H sin² h)
∴ x² = [s² R² (R² — H sin² h)] / [12² R² + s² (R² — H sin² h)]
= R² / [(12² R²) / (s² (R² — H sin² h)) + 1] ∴ x = R / √[(12² R²) / (s² (R² — H sin² h)) + 1] =
H sin δ as given. From H sin δ, the method of obtaining
λ is clear from the formula (H sin λ H sin ω) / R = H sin δ. In
the given numerical example h = 5 nādīs = 360° / 12 = 30°
since 60 nadis of time correspond to 360°.
Thus H sin h = R / 2 ; the remaining work follows.
Verse 86. Another question.
When the Sun is on the prime-vertical the gnomonic
shadow is noted to be 16 inches. The Unnatakāla is 8
nādis. If you could give the H sin δ and s, I shall con-
sider you nothing short of one who is an adept in solving
the totality of the diurnal problems.
Verses 87 and 88. Answer to the question.

317
Here also assume H sin (Unnata) to be the Taddhṛti
as formerly done. Then as the shadow is 16″,
K̇ = √(16² + 12²) = 20″. Then H cos z = (12 R) / K = 12 / 20 × 3438
Unnatakāla = 8 nāḍīs = 48°.
∴ H sin (48°) = assumed Taddhṛti. Then from the
fourth latitudinal triangle, (H sin 48) / (H cos z) = k / 12
∴ Approximate value of k is (12 H sin 48) / (12/20 × 3438) = (20 H sin 48) / 3438 .
This is a known quantity from which s could be computed
since 12² + s² = k². Again from the fifth latitudinal
triangle s / k = (H sin δ) / (S. S.) .
Here s, k and S. S. (Samamandala-Sanku) are known
∴ H sin δ could be got approximately. Thus we have
found approximately the required quantities s and H sin δ.
From this H sin δ and s we have to compute again H cos δ,
Carajyā, Kujyā, etc. and applying the procedure of verse 54
[H sin (Unnata-cāra) × H cos δ] / R + Kujyā = Iṣṭa Hṛti; this
is nearer value of Iṣṭa Hṛti than the assumed Taddhṛti.
From this again obtain as before s and H sin δ; we could
not apply the proportion “ If by the assumed Taddhṛti we
have the previous H sin δ, what shall we have for this
computed Taddhṛti (Iṣṭa Hṛti) ” for the reason given
below. So Repeat the entire process till an invariable
quantity is got which will be the correct value of H sin δ.
Here repeating the entire calculation is correct and
not taking the proportion because Taddhṛti = (R sin δ) / (Sin ϕ cos ϕ)
where sin δ and sin ϕ are both to be computed.
Formerly we could take the proportion in verse 81
because the latitude of Ujjain being known, in the magni-

318
tude of Taddhṛti namely (R H sin δ) / (Sin ϕ cos ϕ) only H sin δ is
variable and Taddhṛti is directly proportional to H sin δ.
But in the present example both H sin δ and H sin ϕ are
both variables so that, that kind of rule of three does not
work.
Verse 89. Oh! Mathematician! At a place where
s = 5'', there 10 nādikas after Sun-rise the shadow S is
observed to be 9''. Tell me what the longitude of the Sun
would be, if you are an adept in computing as well as
understanding the geometry of the sphere.
Verses 90, 91. Answer to the question posed.
Assume H sine (Unnatakāla) to be Iṣṭāntyakā. Then
(K × H cos z × R) / (12 × I. A.) = H cos δ where I. A. = Iṣṭāntyakā.
R² − H cos² δ = H sin² δ ; from this approximate H sin δ
and the given s compute a more approximate I. A. Repeat
the process till an invariable quantity is obtained for
H sin δ, which will be its correct value. From this, using
the formula H sin δ = (H sin λ H sin ω) / R , λ could be had.
Comm. We know the formula for I. A. as
(R³ H cos z) / (H cos φ H cos δ) Assuming Unnatakālajyā as I. A.
H sin (Unnatakāla) = (R³ H cos z) / (H cos φ H cos δ)
∴ H cos δ = (R³ H cos z) / ((12 R. I.A.) / k) = (k × R × H cos z) / (12 × I. A.) ; from
which obtaining H sin δ and proceeding as indicated we
have λ. In the above proof we have used our formula.
But Hindu Astronomers proceed from first principles. Let
us hear Bhāskara. Since S = 9''; K = √(9² + 12²) = 15''

319
∴ Mahā-Sanku = H cos z = (R × 12) / K = (3438 × 12) / 15 = 2750 - 24.
We know that Mahā-Sanku forms a latitudinal triangle
with Iṣṭa Hṛti.
So (H cos z) / (Iṣṭa Hṛti) = 12 / k ∴ Iṣṭa Hṛti = (H cos z × k) / 12
∴ Iṣṭāntyā = (Iṣṭa Hṛti × R) / (H cos δ) ie. H cos δ = (Iṣṭa. Hṛti × R) / Iṣṭāntyā
= (H cos z × k × R) / (12 × I. A.) substituting the above value of Iṣṭa
Hṛti. Here H cos z is got above and I. A. has been assu-
med above as H sin (Unnatakāla).
Note. (1) Computing
H cos δ = (12 R / 15) × (√(s² + 12²) × R) / (12 × H sin (60))
= (R² × 13) / (15 H sin 60) = (13 R × 2) / (15 × √3). Here H cos δ > R which is
invalid.
(2) This is the only place where Bhāskara gave a
numerical example with a slight flaw. In other words,
under the given circumstances the shadow must be greater
than what is given. However, the procedure indicated is
mathematically correct.
(3) It is interesting to note that the flaw was noted
by a commentator named Lakṣmīdāsa as reported by
Munīśwara in his Marīchi Bhāṣya. Munīśwara also
noted the flaw but argues away in an untenable way.
Another commentator named Gaṇeśa who was the author
of the commentary named Śiromaṇipracāśa, does not seem
to have noticed the flaw, or even if he did notice, probably
he fought shy of pronouncing that there was a flaw. In
fact a simple flaw like this in numerical examples, is not
in the least derogatory to the prestige of Bhāskara. So,
the commentators who happened to notice the flaw need
not have pointed the same.

320
(4) Or again in the given place, for the value of
H cos δ to be valid the Unnatakāla x must be such that
15 H sin x > 13 R so that H cos δ might be less than R.
This means sin x > 13/15 = .8667 so that x > 60°—4' ; so
instead of 10 nādīkas, if the time were given to be just
even one Vinādika greater, it would have been alright, or
again if the latitude were given to be just a little less it
would have been alright.
Verse 92. Oh ! Mathematician ! please tell me the
magnitudes of the equinoctial shadow and the longitude
of the Sun if at a place on a particular day, Kujyā is 245
and Taddhṛti 3125.
Verse 93. Answer to the questiou above.
s = √[ 144 Kujyā / (Taddhṛti—Kujyā) ] and H sid δ = (12 Kujyā) / s
and H sin λ = (R H sin δ) / (H sin ω) .
Comm. From the fifth latitudinal triangle compared with third,
Kujyā / Krāntijyā = Krāntijyā / (Taddhṛti — Kujyā) = Agrā / S. S. = s / 12
(1) (2) (3) (4)
Multiplying (1) by (2) Kujyā / (Taddhṛti — Kujyā) = s² / 12²
∴ s = √[ 144 Kujyā / (Taddhṛti — Kujyā) ] Also Equating (1) and (4)
Krāntijyā = (12 / s) Kujyā.
Verse 94. Given that H sin δ + S. S. + Taddhṛti—Kujyā = 6720, and Kujyā + Agrā + H sin δ = 1960. Then
I shall consider him who finds s and the longitude of the
Sun as the very Sun illuminating the lotuses of astronomers.

321
Verse 95. Answer to the question above.
Divide 12 × Second sum by the first sum, that will
be s. Again (12 × Second sum) / (12 + s + k) = H sin δ. From H sin δ, λ
could be had as before.
Comm. Comparing the third and fifth latitudinal
triangles
Kujyā / Krāntijyā = Krāntijyā / (Taddhṛti — Kujyā) = Agrā / S. S. = s / 12
(Kujyā + Krāntijyā + Agrā) / (Krāntijyā + Taddhṛti + S. S. — Kujyā) = 1960 / 6720 = 7 / 24
∴ s = 7/24 × 12 = 7/2 = 3½″.
Again comparing the third and the first latitudinal
triangles
s / Kujyā = 12 / Krāntijyā = k / Agrā = (s + 12 + k) / (Kujyā + Agrā + Krāntijyā)
(1) (2) (3) (4)
= (s + 12 + k) / 1960
(5)
Equating (2) and (5) Krāntijyā = (12 × 1960) / (12 + s + k)
= (12 × 1960) / (7/2 + 24/2 + 25/2) = (12 × 1960) / 28 = 840 since when s = 7/2
k = 25/2 which is the hypotenuse of the triangle formed
by the equinoctial shadow with the gnomon.
Equating (1) and (5) Kujyā = 245 ; equating (3) and
(5) Agrā = 875.
Now from the fourth latitudinal triangle compared
with the first Agrā / s = S. S. / 12 = Taddhṛti / k
(1) (2) (3)
41