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

251
of signs is to be correlated with the modern. We have
from the formula derived out of PZS, A = S + B.
According to modern convention when δ is north, it will be
taken to be positive and when a is to the north of the East
point it also will be taken to be positive so that, (1) when
δ is north and a north, A = S+B ie. B = A-S; this accords
with the Hindu convention namely ‘सौम्येऽन्तरम्’ (2) when
δ is south and a south, - A = S - B so that B = A + S;
this also accords with the Hindu convention, namely
याम्ये योग: (3) But, however, when δ is north and a
south ie. when the Sun having northern declination comes
to the south of the prime-Vertical, A = S - B so that
B = S - A. This accords with the Hindu convention
if only we take B = | A-S | when δ is north.
Bhāskara makes two statements at the end of the
commentary under this verse namely that when δ is south
and A > 2431, there will be no Koṇa-Śaṅku and that
when δ is north and s > 17'' - 5''' there will be four
Koṇa-Śaṅkus. We have to verify these statements. The
first statement is evident because H sin (Agrā) > H sin 45°

252
as may be seen by taking δ = 20°, φ > 61° - 6'. Thus
Bhāskara gave the minimum latitude which could enjoy
four Koṇa-Śaṅkus.
Fig. 41
Fig. 42

253
From fig. 41, it is clear that there are two Koṇa-
Śaṅkus at S₁ and S₂ during the forenoon and similarly
two at S₁' and S₂' in the afternoon where S₁' and S₂'
are the symmetrical points of S₁ and S₂. From fig. 42, it
is clear that if Agrā < H sin 45° when δ is south, there
will be one Koṇa-Śaṅku in the forenoon at S₁ and one in
the afternoon at the symmetrical S₁'.
Verses 31 and 32. H cos z at noon known as
Dinārdha-Śaṅku.
By 'northern hemisphere' it is meant that the Sun
is in the northern hemisphere ie. his Sāyana longitude
ie. modern longitude lies between 0° and 180°, and 'the
southern hemisphere' means that the Sun's longitude lies
between 180° and 360°. The direction of δ may be got
from the above convention. The latitude and colatitude
are always deemed as south and north respectively.
The latitude and colatitude being 'added to subtracted
from or being decreased by' as the case may be, the
declination, we have the zenith-distance and the altitude
of the celestial body at Noon. The zenith-distance and
the altitude are mutually complements.
Comm. In Hindu Astronomy the words “उत्तरगोले”
“दक्षिणगोले” are very often used to connote that the Sun is
on the north or the south of the celestial equator respec-
tively, so that the declination could be automatically known
to be north or south respectively. Regarding the latitude,
the peculiarity in Hindu Astronomy is that what we call
north latitude in modern astronomy is construed as south
in as much as the celestial equator gets depressed south
in northern latitudes. The colatitude SQ in fig. 41 on the
other hand extends north from the south, so that, it is
construed as north.
The word 'Saṃskāra' is used in Hindu Astronomy
in the meaning given above in the translation. Fo

254
example in the equation A = S + B, we say that the
Bhuja is had by a Samskāra between A and S. The
meaning of Samskāra given by Bhāskara is “समदिशोर्योगः
भिन्नदिशोरन्तरम् संस्कारः” Latitude being regarded as southern,
if the Sun's declination is 12° north and the latitude 20°,
then as they are of opposite direction, effecting the
Samskāra as directed 20 - 12 = 8 = zenith-distance
(South) = Nata as it is called similarly 70 + 12 = 82 =
Altitude = Unnata; here we have added because, both
lamba and declination are north. Similarly when δ = 24°
north, and φ = 20° as before (south)
24 - 20 = 4° = zenith-distance (north) = Nata
70 + 24 = 94 = unnata (north). But, we take
180 - 94 = 86°.
In the above working in the first case we found
φ - δ, whereas in the second we found δ - φ. This
difference in treatment is not taken objection to, since,
the word Antara is used to take the positive value of the
difference alone and so in the first instance the nata is
pronounced as south, whereas in the second it is pro-
nounced north.
In modern astronomy, however, we have the formula
z + δ = φ, considering z as positive if south, δ and φ
positive if north. Here 8° + 12° = 20° (first case cited
above) and (- 4°) + 24° = 20° (2nd case, z being negative,
for, it is north. In the Hindu symbolism we have to
pronounce separately when z is south or north, whereas in
modern symbolism the sign alone informs its direction.
Similarly in the equation A = S + B, we have to pro-
nounce ‘north bhuja’ or ‘south bhuja’ as the case may be,
whereas having a convention that δ is + ve when north,
and also the Hindu azimuth (measured from the East
point) the sign of bhuja indicates its direction. In other
words we differentiate the two cases A - S and S - A
giving them signs and deducing the direction of the bhuja

255
from the sign itself without an appeal to a picture or
without ascertaining whether the northern Agrā prevails
over the Southern Saṅku-tala or the Southern Saṅku-tala
prevails over the northern Agrā. Thus the Dinārdha
Saṅku in symbolism = H cos (φ ± d) (20).
Verse 33. Here at noon, Dṛg-jyā is the H sine of
nata and the Saṅku is H sine of unnata.
Second half of 33 and first half of Verse 34.
The product of R and the unmandala-Saṅku divided
by Charajyā is called Yaṣṭi. The Yaṣṭi increased by
Un-mandala-Saṅku gives H cos z according as the Sun is
north or south of the equator.
Comm. Unmandala-Saṅku is H cos z when the Sun
is on the unmandala. From the sixth latitudinal triangle,
wherein Unmandala-Saṅku is Bhuja and Krāntijyā
Karṇa, so by comparing with the second latitudinal
triangle (or rather operating with the second triangle to
signify the Hindu method).
(Krāntijyā × Bhuja) / Karṇa = Unmandala-Saṅku
= (H sin δ × H sin φ) / R (already derived under (19)).
We saw before Charajyā = R tan φ tan δ. Hence as
directed in the verse (R × H sin φ H sin δ) / (R × R tan φ tan δ) = Yaṣṭi
= (H cos φ H cos δ) / R (21).
∴ H cos z (at Noon) = (H cos φ H cos δ) / R ±
(H sin φ H sin δ) / R according as the Sun is on the north or
south of the equator.

256
Hence H cos z (at Noon) = Natajyā
= (H cos φ H cos δ ± H sin φ H sin δ) / R or in modern
symbolism cos (φ ∓ δ) already derived under (20).
We shall now show how the formulation is done by
the simple rule of three (Ref. fig. 39). If a parallel
through o₄d is drawn to cut Aa at a₁, then Aa₁ is called the
Yaṣṭi, which is vertical. The triangles Aoa₁ and Do₄d
are similar so that Aa₁ / Ao = Dd / Do₄ ∴ Aa₁ = (DD × Ao) / Do₄ .
But Ao / Do₄ = R / Charajyā for, all the lines of the diurnal
circle and the equator stand in the ratio
(H cos δ) / R = Ao / R = Do₄ / Charajyā = Kujyā / Charajyā ∴ Ao / Do₄ = R / Charajyā
∴ Aa₁ = (Unmandala Śaṅku × R) / Charajyā = Yaṣṭi as
formulated.
Now Dinardha Śaṅku = Aa = Aa₁ + a₁ a = Aa₁ + Dd
= Yaṣṭi + Unmandala Śaṅku. It is evident from fig. 21
why in the northern sphere the sum is to be taken
whereas in the southern, the difference is to be taken.
Latter half of verse 34. Definition of Hṛti and
Antyā.
The sum or difference of Dyujyā and Kujya will be
similarly Hṛti, whereas the sum or difference of Charajyā
and radius will be Antyā.
Comm. We defined formerly Hṛti and Antyā under
our commentary on the latitudinal triangles. From fig. 39
Hṛti = o₁A = o₁o + oA = o₄D + oA = Kujya + H cos δ
(Dyujyā) (22).

257
In the parallel great circle, the Equator we have
therefore Antyā = Charajyā + R (already derived).
Verse 35. Antyā = (Hṛti × R) / (H cos δ) = (Hṛti × Charajyā) / Kujyā and
∴ Hṛti = (Antyā × H cos δ) / R = (Antyā × Kujyā) / Charajyā by what is
called Guṇa-ccheda-Viparyaya ie. alternando.
Comm. Evident.
Verse 36. To obtain Dinārdha-Śaṅku from Antyā
and Hṛti
(Antyā × un-maṇḍala Śaṅku) / Charajyā = (Hṛti × 12) / K = Dinārdha-
Śaṅku.
Comm. The second formula is derived from the
similarity of △ Ao₄a (fig.) with the first latitudinal triangle.
From the similarity of Aao₄, Ddo₄ fig. 39,
Ao₄/Do₄ = Aa / Dd ∴ Aa = (Hṛti × unmaṇḍala-Śaṅku) / Kujyā .
But Aa = Dinārdha-Śaṅku
∴ Dinārdha Śaṅku = (Hṛti × U.S.) / Kujyā (U. S. = Un-
maṇḍala-Śaṅku.)
But Hṛti / Kujyā = Antyā / Charajyā
∴ D.S. (Dinārdha-Śaṅku) = Charajyā / Antyā × U.S.
= (Hṛti × 12) / K VIII.
Verse 37. Meridian Zenith distance.
Agrā ± (Hṛti × bhuja of a lat. triangle) / (Karṇa of a lat. triangle) = H sin z
where z is the meridian zenith distance.
33

258
Comm. This formula is a special case of the formula
A = S + B since the H sine of the meridian zenith
distance is the Śaṅku-Bhuja at noon. From fig. 39
(Hṛti × bhuja of a latitudinal triangle) / (Karṇa of a latitudinal triangle) = O₁a = Dinārdha
Śaṅkutala.
The operation of sign has been already explained.
Verse 38. An alternative method.
The meridian zenith distance of the Sun can be had
also by the formula (Hṛti ± Taddhṛti) B.L.T. / K.L.T. where
B.L.T. and K.L.T. are the bhuja and karṇa of any
latitudinal triangle.
Comm. (Ref. figures 43 and 21). Let E₁ be the
centre of the armillary sphere so that QE₁R is the
diameter of the celestial equator which is on the median
plane. Let S₁ F₁ S be the diameter of the diurnal circle
of the Sun, which is also on the meridian plane so that
S₁ F₁ is the Taddhṛti, S₁ S is the Hṛti and S the position
of the Sun on the meridian.
H sin z = SM = SF₁ sin F̂₁ = SF₁ sin ϕ =
(Hṛti − Taddhṛti) × s/K where s/K can be replaced by
B.L.T. / K.L.T. (s=equinoctical shadow and k the Viṣuvat-Karṇa).
In the Southern sphere, H sin z = S′N = S′F₁′ sin ϕ
= (S′S₂ + S₂F₁′) sin ϕ = (Hṛti + Taddhṛti) × s/k.
Verse 39. Still another way of obtaining the m. z. d.
(meridian-zenith-distance).
R − H versin (altitude) = H sin z

259
Fig. 43
This formula gives not only the meridian zenith
distance but H sine of the zenith-distances of the Sun
when he is on the Koṇa-Vṛtta or prime-vertical or
unmaṇḍala.
Comm. From fig. 43, H sin z = SM = LE₁ = E₁s −
sL = R − H versine (Ss) = R − H versine (altitude) as
given. Since R − H versine (altitude) = R − {R −
H cos (90 − z)} = R − (R − R sin z) = H sin z, so this
formula applies wherever the Sun be. This is almost
begging the question as H sine of z is being sought
through H versine of (90 − z).

260
First half of the verse 40. To obtain the shadow S
and K the Chāyakarṇa of any shadow
(H sin z × 12) / (H cos z) = S and (R × 12) / (H cos z) = K.
Comm. (Ref. fig. 44).
(12. H sin z) / (H cos z) = 12 tan z = S. Also 12 / K = cos z = (H cos z) / R
so that (12 R) / (H cos z) = K.
The Hindu method of looking at this through the
similarity of ΔS OM☉ and Ogn the gnomonic triangle.
is as follows. ☉M is called Mahā-Śaṅku ie. H cos z ; ☉L
is Dṛkjya or
H sin z = OM. 12 / (H cos z) = S / (H sin z) so that S =
12 H sin z/H cos z. Also,
On / O☉ = 12 / (H cos z) ie. K / R = 12 / (H cos z) ∴ K = (12 R) / (H cos z) .
It will be noted that fig. 44 pertains to any vertical plane.
Second half of Verse 40. The Dinārdha-Karṇa is
equal to (R × k) / Hṛti where k is the Viṣuvat-Karṇa.
Comm. The formula is derived through twice apply-
ing the rule of three or what is the same, through the
similarities of two sets of triangles
From fig. 39, O₁A / Aa = Hṛti / Dinārdha-Śaṅku = k / 12 (a)
and from fig. 44 On / O☉ = 12 / Dinārdha-Śaṅku = K / R (b)
where K is the required Chāyākarṇa.
Dividing (a) by (b)
Hṛti / 12 = k / 12 × R / K ∴ K = kR / Hṛti

261
Fig. 44
Verse 41. Alternate method of obtaining K
101530/H sin λ = para (say) where λ is the Sāyana
longitude of the Sun ; then, (Para × k) / s = K where K is
un-mandala-Karṇa
First half of Verse 42. To obtain K when the Sun is
on the prime-vertical— Para × s/k = Samavṛttakarṇa.
Comm. From fig. 19, from the similarity of triangles
BD ☉ and CMA, B ☉ / CA = ☉ D / AM ie. ☉ D = (B ☉ × AM) / CA ie.
H sin δ = (H sin λ H sin ω) / R (a)
Then consider the similarity of the first and the sixth
latitudinal triangles ; then Unmandala Sanku / Krāntijyā = s / k (b)
where s is the equinoctial shadow and k the Viṣuvat-
Karṇa. Again taking that ☉ the Sun lies on the unman-

262
dala in figure 44, (H cos z) / 12 = R / K = (Unmandala Śaṅku) / 12 (c)
Eliminating Krāntijyā and Unmandala Śaṅku from
(a), (b) and (c) (Unmandala Śaṅku) / (H sin λ H nis ω/R) = s / k
∴ (12R / K) / (H sin λ H sin ω / R) = s / k ie. 12R² / (KH sin λ H sin ω) = s / k
∴ K = (12R² × k) / (s H sin λ H sin ω). Here 12R² / (H sin ω) = (12 × 3438²) / 1397
= 101531 ; but Bhāskara has taken 101530 taking a
more correct value of R. Then 101530 / (H sin λ) is symbolized as
para so that para × k / s = K = Unmandala Karṇa. Regard-
ing the Samavṛttakarṇa, in the place of (b) above we have
(Sama-Śaṅku) / Krāntijyā = k / s (b') by the similarity between the
first and the fifth latitudinal triangles. Equation (c)
holds good with respect to any H cos z and the corres-
ponding K since 12 R = K × Śaṅku and 12 R is a cons-
tant. Noting therefore R / K' = (Sama-Śaṅku) / 12 (c')
eliminating Krāntijyā and Sama-Śaṅku among (a), (b'),
(c'), we shall have
K = (12 R² s) / (k H sin λ H sin ω) = para × s / k as stated.
Second half of Verse 42. To obtain the Dinārdha-
karṇa from the Unmandalakarṇa.
(Un-mandalakarṇa × Charajyā) / Antyā = Dinārdhakarṇa.
Comm. We have equation (c) above stating
12 R = K × Śaṅku. (c) But

263
Iṣṭa Śaṅku / Iṣṭa Hṛti = cos φ = constant = Dinārdha Śaṅku / Hṛti
= Sama-Śaṅku / Taddhṛti = Unmaṇḍala Śaṅku / Kujyā (d) (23)
Again by virtue of the proportionality of
Iṣṭa Hṛti / Iṣṭāntyā = Hṛti / Antyā = Kujyā / Charajyā (e) (24)
We have Ishta-Śaṅku / Ishtāntyā = Dinārdha Śaṅku / Antyā
= Unmaṇḍala Śaṅku / Charajyā
∴ Iṣṭa Karṇa × Iṣṭāntyā = Dinārdha Karṇa × Antyā
= Unmaṇḍala Karṇa × Charajyā (f) (25)
∴ Dinārdha Karṇa = (Unmaṇḍala Karṇa × Charajyā) / Antyā
as stated in the verse.
Verse 43. (Unmaṇḍala Karṇa × Kṣitijyā) / Hṛti
= (Sama Vṛtta Karṇa × Taddhṛti) / Hṛti = Dinārdha Karṇa.
Comm. From (c) and (d) above Dinārdha Karṇa ×
Hṛti = Sama Karṇa × Taddhṛti = Unmaṇḍala Karṇa ×
Kujyā (g). Khitijyā is the same as Kujyā.
From this the statement follows :
Verse 44. The ancient Achāryas found the gnomonic
shadows when the Sun is on the meridian, prime-vertical
and the Kona-Vṛtta (ie, Vertical when the northern or
southern Hindu azimuths are 45°) by different methods.
I consider him to be the very Sun illuminating the lotus-
faces of aitronomers, if anybody could give a method to
find the shadow in any required direction, which holds
good in all cases universally.

264
Comm. Evident.
Verse 45. Definition of Dikjyā H sin (azimuth). The
angle between any vertical and the Prime-Vertical
measured on the horizon is what is called Digamsa and its
H́ sine is known as Dik-jyā either in the Eastern hemi-
sphere or the Western.
Comm. In modern astronomy azimuth is measured
along the horizon from the north point towards the east
point round the horizon. In Hindu Astronomy however,
the azimuth is measured from the East point on either
side and from the West point also on either side specifying
whether it is north or south.
Verse 46 and first half of 47. To obtain the
gnomonic shadow in any arbitrary direction.
Assume Rs / (H sin a) as the equinoctial shadow and
obtain the H sine of the corresponding latitude L. Then
the product of that H sin L and H sin δ divided by
H sin ϕ will give H sin D where D is a hypothetical decli-
nation. With the new L and this D, as the hypothetical
latitude and declination, obtain the meridian zenith
distance by the formula Z + D = ϕ, and through this
m.z d. obtain the shadow, which will be the shadow in the
required direction namely 12 tan (ϕ ± D).
Comm. Let gL be the gnomonic shadow on the
equinoctial day in a given direction given by a°
Digamsa (the Hindu azimuth) and let gN be the shadow in
the same direction on any day. (fig. 45) We know that the
extremity of the gnomonic shadow on the equinoctial day
traces a straight line parallel to the East-West line Eω
at a distance of the equinoctial shadow s because the
Equatorial plane passing through the foot of the gnomon

265
Fig. 45
and that passing through the top of the gnomon being
parallel planes cut the horizontal plane in parallel straight
lines. (This will be also proved analytically subsequently).
Hence LM = s. Now from the figure
LM / gL = AB / gA = H sin a / R ∴ gL = H sin a / Rs I
This gL is spoken of as Iṣṭa-Drikmandala palabhā
because it is the shadow on the equinoctial day in
any vertical. LN is the increment in the shadow on
account of declination and we have to compute this and
correlate gL and LN. For this refer to figs. 46 and 47.
In fig. 46, QRT is the equator, so that when the Sun is on
the equator on the equinoctial day in the direction given
by ZS, ZT is the zenith-distance. Let ZS be the zenith-
distance of the Sun in the same direction on any day
From the analogy of finding H sin δ from H sin λ, from
this figure
34

266
H sin SR = (H sin ST × H sin T̂) / R II and
H sin φ = (H sin ZT × H sin T̂) / R III so that
(H sin SR) / (H sin φ) = (H sin ST) / (H sin ZT) ∴ H sin ST = (H sin SR / H sin φ) × H sin ZT.
Noting that SR = δ and putting ST = D
H sin D = (H sin δ / H sin ϕ) × H sin ZT.
[Diagrams: Fig. 46 and Fig. 47 showing spherical triangles with vertices P, Q, N, R, S, T, D, δ]
Fig. 46 Fig. 47
The same formulae are derivable from fig. 47 also; only
in fig. 46 while there is a decrement in the shadow of the
day as compared with the shadow on the equinoctial day,
in fig. 47, there is an increment. This is seen from the
decrease and increase of ST in the zenith-distance ZT of
the equinoctial day in the given direction. Now corre-
lating fig. 45 with figures 46 and 47, the shadow gL
pertains to the zenith-distance ZT on the equinoctial day
whereas the shadows gN pertains to the zenith-distance on
the day concerned in the same direction. We have,
S / √(12² + S²) = (H sin z) / R so that RS / √(12² + S²) = H sin z where
S is the shadow at any instant when the zenith-distance
is z. The process indicated by saying ‘Obtain H sin φ

267
construing Rs / (H sin a) as the equinoctial shadow', means
computing H sin ZT and there from ZT from the shadow
gL of fig. 45. Then the process indicated by saying
“Obtain H sin D = (H sin ZT × H sin δ) / (H sin φ) and therefrom
D'' means computing ST.
Then clearly ZS = ZT ± ST, ie. the required zenith-
distance is got by what is technically called Samskāra
between ZT and ST as is stipulated between φ and δ to
obtain Z from the formula Z ± δ = φ (The word
Samskāra was defined as meaning addition when the
directions are the same and difference when they are
opposite). Then the gnomonic shadow is got from this
zenith-distance using the formula S = (12 H sin z) / (H cos z) .
Thus the procedure adopted by Bhāskara was con-
ceived by him first having Fig. 45 before him and then
using figures 46 and 47. In this particular process,
H sin a is given and H sin δ also, which means that it
is sought to find the shadow on a given day in a given
direction.
Incidentally we shall find the locus of the extremity
of the gnomonic shadow during the course of a day. Let
in fig. 48 g represent the gnomon's foot, and S the shadow
whose extremity is p. Required to find the locus of p.
Take the gnomon to be of unit length so that the length of
the shadow S = 12 tan z becomes tan z here. Take Eω
and sn the east-west line and the north-south as the
axes. Then we have
x² + y² = tan² z (1). But we have from the triangle
PZS sin δ = sin φ cos z + cos φ sin z sin a
ie. sin δ / (cos φ cos z) = tan φ + tan z sin a = tan φ + y (2)
ie. sec z / A = y + tan φ when A = cos φ / sin δ .

268
Fig. 48
But sec z = √(1 + tan² z) = √(x² + y² + 1)
∴ √(x² + y² + 1) / A = y + tan Φ which reduces to
x² + y² (1 - A²) - 2A² y tan Φ + 1 - A² tan² Φ = 0 (3)
From this it is evident that the locus is an ellipse or
parabola or hyperbola according as A ⋚ 1; also it will
be seen that the eccentricity is A. The locus is wrongly
stated to be always a hyperbola in some text books. For
it to be an ellipse A < 1 ie. cos Φ < sin δ ie. δ > 90 - Φ
ie. Φ + δ > 90. In such latitudes and under such decli-
nations, it will be an ellipse ie. at a place just north of
the place where the perpetual day just begins the locus will
be an ellipse. Hence in the arctic region it will be always
an ellipse; and in the place just at which the perpetual
day begins it will be a parabola and in the lower latitudes
it will be a hyperbola, ie. it will be a parabola where the
latitude Φ is given by 90 - δ.
When Φ = 90°, A = cos Φ / sin δ = 0 provided δ ≠ 0. If,
however, in addition δ = 0, A becomes indeterminate, but
we may note then, that the Sun will be circling round the
horizon on that equinoctial day at the north pole. We

269
may further note that at the north pole, the altitude of
the Sun is always δ so that the length of the shadow cast is
always equal to cot δ and this will be infinite when δ = 0.
If now φ = 90°, and δ ≠ 0, though A = cos φ / sin δ be
comes zero, A tan φ will not be zero because A tan φ =
cos φ / sin δ × tan φ = sin φ / sin δ = 1 / sin δ (∵ φ = 90°). On the
other hand A² tan φ = A × A tan φ = / sin δ = 0 because
A = 0. Thus the term containing y in eqn. (3) vanishes.
∴ The equation reduces to
x² + y² = A² tan² φ - 1 = cosec² δ - 1 = cot² δ (ie. +ve)
ie. at the north pole the locus will be a circle with radius
cot δ. When
A = ∞ the locus √(x² + y² + 1) / A = y + tan φ becomes
y = - tan φ which means that the hyperbola degenerates
into the straight line which is parallel to the east-west
line and is in the north at a distance of tan φ ie. s, the
equinoctial shadow since the length of the gnomon
is taken to be unity. In particular when A tan φ = 1 ie.
φ = δ, the constant in (3) is zero, so that the locus
passes through the foot of the gnomon as is also evident
from the fact that the Sun passes through the zenith.
Taking a northern latitude say 17°, the loci of the
extremity of the shadow are shown in fig. 48A (page 270)
on important days when δ=ω, when δ=0, when δ = - ω,
when δ = φ. The maximum mid-day shadow is
tan (φ + ω), when δ = - ω taking the gnomon's length
to be unity; this shadow is cast north of the gnomon along
the south-north line through the gnomon, on Dec. 23rd of
the year. The minimum length of the mid-day shadow
occurs when δ = φ, the shadow being zero and being at the
foot of the gnomon, the Sun being then just overhead. The
maximum shadow cast south of the gnomon at mid-day
is tan (ω - φ).

270
Shadow on Dec. 23
No Shadow in this direction on Dec 23
Immediately before 21st March
B C A On 21st March
Immediately after 21st March
No Shadow in this direction on 22nd June
W B C A E
JUNE when δ=φ
On 22nd JUNE
Fig. 48A Showing the locus of the extremity of the shadow on different
days at a latitude of 17°
Note. OA, OB are the shadows computed by Bhāskara when the Sun is on
the prime-vertical.
In the method of finding the shadow under verse 46,
we perceive Bhāskara's genius in (1) looking upon the
shadow as being made up of two segments namely that due
to φ and that due to the declination (2) in conceiving
what he calls Iṣṭa-drik-mandala palabhā, and Iṣṭa-drik-
mandala Krānti and (3) in deriving the equation
H sin D = (H sin ZT × H sin δ) / (H sin φ)
Latter half of verse 47 and verse 48. Something
to be noted.
In computing the shadow in a given direction, there
may be two shadows at times in the northern hemi-
sphere. When H sin a < Agrā and there will be none in
the southern. To compute the second shadow we have to
take 180 − L also as the latitude where L is the latitude
computed, and proceed in the same way as we have done
before.