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

157
Moon's hourly motion is given in modern almanacs. Since
it is very cumbrous to use the method of successive approxi-
mation to determine the ending moment of every tithi, the
Hindu almanac-makers generally compute the ending
moment of a tithi using the daily motion of the moon
computed for the moment of Sun-rise of the day. Only for
ritual purposes, the method of successive approximation is
used and also in the computation of eclipses.
Verse 39. Computation of the Sīghragatiphala.
δl − [H sin (90−E₂) × δm] / K = δS where δl is the daily
motion of the Sīghrōcca, E₂ = Sīghraphala, δm = daily
motion in the Sīghra mean anomaly, K = Sīghrakarṇa,
and δS the true motion of the planet. If δS is negative the
planet is retrograde.
Comm. This is mathematically an important verse,
and the proof given by Bhāskara really reflects his
genius. Before we attend to his proof, we shall give this a
modern treatment. (Ref. figs. 14, 15). SA and EA' are the
Fig. 14
heliocentric and geocentric directions to Aswini the Hindu
Zero-point of the Zodiac; S = Sun, E = Earth; V = In-
ferior planet Venus or Mercury; J = Superior planet; E =
Sīghra-phala, K = Sīghra Karṇa; M = Sīghra anomaly.

188
[चित्र: Fig. 15]
Fig. 15
True motion of the planets = δ (A'EV) or δ (A'EJ)
But δ (A'EV) = δ (A'EV'–n̂) and δ (A'EJ) = δ (A'ES–n̂)
In the case of the Inferior planet δ (A'EV') = δ (ASV)
= Sīghrōcchagati and δn is Sphutakendragati where n is
called Sphutakendra, m being called Madhyakendra. In
the case of the Superior planet δ (A'ES) = S'īghrōcchagati
because the Sun plays the part of S'īghrōccha in the case of
a Superior planet and δn = Sphutakendragati as before.
Hence in both the cases, Sphutagati = S'īghragati – Sphuta-
kendragati. We have now to find Sphutakendragati to
obtain Sphutagati, as Bhāskara remarks rightly “महामति-
द्भिः केन्द्रगतिरेव स्पष्टीकृता”. In other words we have to find
δn. From the figures K cosn – R cos m = r (i) Differentia-
ting this we have – K sin n δn + cos n δK + R sin m δm
= 0 (ii). But K² = R² + r² + 2Rr cos m so that 2δK × K
= – 2Rr sin mδ m (3) Eliminating δK between (2) and
(3) – K sin nδ n – (Rr sin mδ m × cos n) / K + R sin mδm = 0
i.e. K sin nδn = R sin mδm ( 1 – r / K cos n )
= (R sin mδm / K) (K – r cos n)

159
But K — r cos n = R cos E.
∴ δn = (R sin mδm × R cos E) / (K² sin n) But R sin m = K sin n
Cancelling δn = (R cos Eδm) / K = (H cos E δm) / K as given by
Bhāskara. In the formula H sin E = (r H sin m) / K, Bhāskara
perceived the variability of both H sin m and K on
the right hand side and he exclaims “ न हि केन्द्रगतिजमेव
फलयोरन्तरं स्यात्, किन्त्वन्यथाऽपि अद्यतनभुजफलश्वस्तनभुजफलान्तरे
त्रिज्यागुणे अद्यतनकर्णहृते यादृशं फलं न तादृशं श्वस्तनकर्णहृते, स्वल्पा-
न्तरेऽपि कर्णे भाज्यस्य बहुत्वात् बह्वन्तरं स्यादित्येतदानयनं हित्वा अन्यत्
महामतिमद्भिः कल्पितम्, तद्यथा केन्द्रगतिरेव स्पष्टीकृता ” i.e. “ The
variation in the Sīghraphala is not entirely constituted
by the variation in m but also by that in K......So
leaving the method of seeking δE through the formula
H sin E = (r H sin m) / K the great intelligent astronomers
used the formula.
Sphutabhukti = Sīghra Bhukti — Sphuta Kendra-
bhukti (Bhukti means gati) wherein it was'sought to obtain
the variation in Sphutakendra i.e. n̂ in the figures.
We have given a proof of Bhāskara’s formula, which
circumvented finding Sīghragatiphala but which sought
directly Sphutabhukti, by using the modern heliocentric
figures. We shall now see how Bhāskara could deal with
such a tough problem. Refer fig 16. Let P₁, P₂ be two
positions of the planet on two consecutive days relative to
the Sīghra A₂, so that P₁E₁P₂ is the Sphuta Kendragati
spoken of. It will be noted that it is not Sphutagati be-
cause P₁, P₂ are positions of the planet relative to A, which
is itself moving (as rightly remarked by Bhāskara). It is
Sphuta Kendragati because A₂E₁P₁ and A₂E₁P₂ are the

160
Kendras on two consecutive
days whereas Madhyakendragati
^
is P₁E₂P₂. Also, we have the
equation.
Sīghra — Sphutagraha =
Sphutakendra so that Sīghra-
gati — Sphutagati = Sphuta-
kendragati. Hence Sphutagati
= Sīghragati — Sphutakendra-
Fig. 16 gati. So we have now to seek
the value of P₁Ê₂P₂. Let P₁a stand for the Sīghra-
phala of the first day which is equal to e f, f being the true
planet of the first day. E₂ d will be parallel to E₁ f because
P₁ e being parallel to E₁ E₂ and e f being equal to P₁ d, d f
11 P₁ e (parallels to E₁ E₂ cut off equal arcs on the two
circles). This may be seen also as follows. Since P₁ d is
taken to be equal to e f, e being the mean planet and f the
true on the first day e Ê₁ f = P₁ Ê₂ d. But E₁ e 11 E₂ P₁
∴ e E₁ f = E₁ P̂₁ E₂ ∴ P₁ Ê₂ d = E₁ P̂₁ E₂ and alter-
nate angles being equal E₁ P₁ 11 E₂d. P₁ a is the H sine
of P₁ d where P₂ b is the H sine P₂ d. Looking upon P₁P₂
as an increment in P₁ d i.e. looking upon the Kendragati
P₁ Ê₂ P₂ as an increment in Sīghraphala, Bhāskara uses
the method of Bhogyakhanda sphuti Karana to obtain the
Sphutakendragati. From the figure.
P₂c = P₂b — P₁a = H sin (G + δm) — H sin E
H sin E H cos δm + H cos E H sin δm
= ————————————————————————————————————— — H sin E
R

161
Taking H cos δ M = R and H sin δm = δm
P₂c = (H cos E δm) / R This is at the end of E₁ P₂ i.e.
at the end of K; so to get the corresponding chord in
the deferent we do Karnānupāta so that the result is
(H cos E δm) / R × R / K = (H cos E δm) / K as given by Bhāskara.
We have cut short Bhāskara’s method of Bhōgyakhanda
Sphutīkaraṇa to make it clear to a modern student. Since
P₂c is small, passing on from the H sine to the arc is not
necessary, for, the H sine of a small arc is equal to the arc
itself.
Bhāskara’s argument, however, is as follows:— “If for
225, we have Bhōgya Khanda, what for δm?” The result
is (B × δm) / 225; Then B is rectified as follows:— “When the
H cosine E is equal to the radius, i.e. initially in the H sine
table, the Bhōgyakhanda is 225, then what is it for H cos E?”
The result is (H cos E × 225) / R which we have to substitute
for B. Then if this be at the end of K what is it at
the end of R?” The result is (H cos E × 225) / R × δm / 225 ×
R / K = (H cos E × δm) / K as given.
Before we proceed to explain ‘शेषं च वक्रा विपरीतशुद्धौ’,
we shall explain what Bhāskara pointed out as a mistake in
Lallācharya.
Verse 40. Let Mathematicians understand that what
formula was given by Lallācharya for Sīghragatiphala is
not correct. When the anomaly is 90° or 270°, the gati-
phala vanishes and there will be gatiphala at the points
where it ought to be Zero according to his formula.
21

162
Comm. Ref. verse 45, Spaṣṭādhikāra, Śiṣya Dhī-
vṛddhida तद्रहिताऽऽशुभुक्तिः, त्रिज्ज्याहता स्वचलकर्णहताऽऽशुचापभोग्य
ज्यया विगुणिता विहृताऽऽद्यमौर्व्या, लब्धं त्यजेत्, स्वचलतुङ्गगतेः सदैव शेषं
स्फुटा भवति च ग्रहभुक्तिरेवम्" i.e.
( शीघ्रगति-मन्दस्फुटगति ) × R Śīghraphala Bhōgyakanda
--------------------------- × -----------------------
K 225
= Sīghragatiphala.
Here the quantity within the brackets is δm. What
Lallācharya had in his mind is as follows. ‘If for the Ādya
Khanda 225 we have H cos E = R, what shall we have for
the Bhōgya Khanda of the Sīghraphala ?’ The result
would be [ H sin (E + 225) - H sin E ] × R / 225 =
( H sin E H cos 225 + H cos E H sin 225 / R - H sin E ) × R / 225
Taking H cos 225 = R and H sin 225 = 225 it would be
H cos E × 225 / R × R / 225 = H cos E. So Lallācharya’s
formula would become (δm / K) × H cos E as given by Bhās-
kara. The charge levelled at Lallācharya is due to the fact
that by the word ‘Āsu-chāpa’ by which Lallācharya meant
‘आशुफलचाप’ Bhāskara meant ‘आशुकेन्द्रचाप’. When the
Kendra = 90 or 270, the Bhōgyakhanda being Zero, the
Sīghragatiphala would be Zero. Also where it ought to be
zero namely (90 + H sin⁻¹r) & (270 - H sin⁻¹r) it would not
be zero. If Lallācharya had really meant what Bhāskara
allributed to him, the formula would be (H cos mδm / K) and
it is very unlikely that Lallācharya would have meant this
wrong formula, for, even if K were taken by him to be
steady, δ (r H sin m / K) = (r H cos mδm / K) and Lallāchar-
ya’s formula does not contain ‘r’. The only non-rigorous
part in Lallācharya’s formula is at the point where he took

163
H̄ cos 225 = R and H sin 225 = 225 which is rather crude.
Bhāskara of course improved on this crudeness by taking
δm to be an increment in E.
In the commentary under this verse, Bhāskara,
having misinterpreted Lallācharya's phrase आशुचाप as
meaning आशुकेन्द्रचाप and not as आशुफलचाप which was
in the mind of Lallācharya, goes on recounting examples
where the wrong formula attributed to him would give
wrong results. So, we need not enter into those details.
Now we shall correlate Bhāskara's formula with its
modern counterpart. Assuming coplanar heliocentric cir-
cular orbits for planets, let us see at what points two planets
appear mutually stationary, that is, have a Zero relative
angular velocity before they appear mutually retrograde.
Let S = Sun, E = Earth, J = Jupiter, u = Earth's
linear velocity, v = Jupi-
ter's linear velocity r and
R the orbital radii of the
Earth and Jupiter respe-
ctively. Let EE' and JJ'
be perpendiculars to EJ so
that when the relative velo-
city of Jupiter with respect
to the Earth perpendicular
to EJ is Zero, Jupiter will
appear stationary as seen
from the Earth. This
Fig. 17
means that u cos θ + v cos ξ = 0 I
∴ u / v = - cos ξ / cos θ II
But from triangle ESJ,
R cos ϕ + K cos θ = r III and
r cos ϕ + K cos ξ = R IV

164
From III & IV cos ξ / cos θ = (r cos ϕ − R) / (R cos ϕ − r) V
Equating cos ξ / cos θ from II and V
− u / v = (r cos ϕ − R) / (R cos ϕ − r) so that (ru + Rv) / (rv + Ru) = cos ϕ
If m be the Śīghra anomaly m = 180 − ϕ
so that cos m = − (ru + Rv) / (rv + Ru) VI
Now we shall show that Bhāskara's formula accords with
this Spaṣṭagati = Śīghragati − (H cos E δm) / K . As per
Bhāskara
Spaṣṭagati = 0 if Śīghragati = (H cos E δm) / K VII
ie. Jupiter appears stationary as seen from Earth, if Śīghra-
gati = (H cos E δm) / K
The angular velocities of Earth and Jupiter are respect-
ively u / r and v / R so that the Sun's apparent velocity is also
u / r and δm = Kēndra gati = Sun's apparent velocity
minus Jupiter's heliocentric velocity = u / r − v / R
∴ Substituting in VII
Śīghragati = u / r = (H cos E / K) (u / r − v / R)
∴ u / r ( (H cos E / K) − 1 ) = (H cos E / K) × v / R
ie. u / r ( (H cos E − K) / K ) = (H cos E / K) × v / R

M₂ + E₂ = S I where M₂ = Mandasphuta-
graha and E₂ = the Sīghraphala where from we have
δM₂ + δE₂ = δS II i.e. Mandasphutagati + Sīghragati
phala = Spastagati (a) Let E₂ be maximum so that
δE₂ = 0, then δM₂ = δS. This means that in the helio-
centric figures 11 and 12, at the points a and b,: the
Sīghraphala being maximum, the Mandasphuta gati will be
itself the Spashtagati. This line ab, it will be seen corres-
ponds with the so called “कक्षामध्यगततिर्यग्रेखा (प्रतिवृत्तसम्पातरेखा)”
i.e. the line cutting the eccentric, drawn through the centre
of the deferent.
(b) The planet begins to retrograde only after the
Spashtagati vanishes i.e. after δM₂ + δE₂ becomes Zero.
Taking δM₂ to be almost a constant since the Mand

166
negative when the planet courses the arc of the smaller
segment ab, because E₂ decreases from a maximum value
to zero. As a matter of fact δE₂ negatively increases along
ac and again increases from a negative minum at c to Zero
at b as we course along cb (vide fig. 11 & 12). Thus the
planet will assume zero velocity at two points symmetrical
about c and in between ab. In other words the planet will
be retrograde along the arc d₁ cd₂ not entirely along ab as
some have misconstrued. Regarding E₂, it is clear that it
is zero at S'. then gradually increases to a maximum as the
planet traces S'a, the maximum being assumed at a, then
it decreases from the maximum to zero as the planet
courses the arc ac. and then increases from zero to a maxi-
mum at b and finally decreases from that maximum to
zero at S' again. Keeping the Earth constant it will be
noted that an Inferior planet always goes anticlockwise
whereas a Superior planet always goes clockwise. Also
it will be seen that the Sīghraphala is positive as the planet
courses the arc Sac, whereas it is negative as it courses
cbS'.
(c) The values of the spashtagati at S' and C will be
respectively putting H cos E = R in the formula.
Spashtagati = Sīghragati — (H cos Eδm) / K . Here for
Sīghragati we may put U and U — V for δm, and putting
K = R + r at S' and R — r at C, the values of the spash-
tagati would be
[U — R (U — V)] / (R + r) and [U — R (U — V)] / (R — r) ie. (RV + ru) / (R + r),
(RV — rU) / (R — r) respectively.
(RV + rU) / (R + r) > (RV — rU) / (R — r) if R²V — rRV + RrU — r²U >
R²V — RrU + rRV — r²U

167
ie. if rR (U - V) > Rr (V - U). U - V is + ve and
equal to V - U.
So the positive velocity at S' of the planet will be
equal to its negative or retrograde velocity at C. A quan-
tity assuming values K, O, - K,O,K must have a value
numerically less than K in between. Thus the velocity
direct or retrograde at any point of the orbit is less than
the numerical value of the velocity at S' and C.
Verse 41. Retrograde motion.
The planets Mars, Mercury, Jupiter, Venus and Saturn
will be retrograde when the Sīghra anomaly assumes values
163, 145, 125, 165 and 113 respectively and the direct
motion again ensues at (360 - 163), (360 - 145), (360 - 125),
360 - (165) and (360 - 113) respectively.
Comm. Since we have had the formula VI
cos m = - ( (rU + Rv) / (rV + RU) ) for stationary points,
noting that cos (180 - θ) = - cos θ = cos (180 + θ) the
stationary points are symmetrically situated with respect
to S' of figures 11 and 12. As mentioned before substitut-
ing the values of U, V, r, R for all the planets we can
prove the veracity of Bhāskara's statement.
Verse 42. Heliacal rising and setting of planets.
Mars rises heliacally in the East by 28°, Jupiter by
14°, Saturn by 17° of Sīghra anomaly and set heliacally in
the west by degrees which are the differences of the above
and 360° respectively.
Comm. The Sun's velocity being greater than that of
the Superior planets, the Sun overtakes them so that they
set in west and rise in the East. When these planets are

168
situated within particular limits from the Sun, they will be
invisible in the rays of the Sun. As these superior planets
will be near the Sun, near the moment of conjunction,
they will not be seen at conjunction and within particular
limits from the position of the Sun. The limits cited
above mostly depend upon their respective brilliance and
to some extent upon their distances from the Sun too.
Their brilliance again depends upon the extent of their
gibbosity. The formula given for the phase in modern
astronomy is phase = (1 + cos EpS) / 2 and noting that Êps =
E₂ = the Śīghraphala, phase = (1 + cos E₂) / 2. Hence the Supe-
rior planets are always gibbous ie. the disc illuminated will
be always greater than ½. Though at conjunction E₂ = 0
and the entire discs of the major planets will be illumina-
ted, we cannot see them as they are immersed in the rays
of the Sun. As they emerge out of conjunction gradually
E₂ will be increasing so that the discs will be illuminated
lesser and lesser gradually. But K decreasing, the brilli-
ance will not be so much effected. Along the arc acb (figs.
11, 12) the planets gradually gain in illumination and will
be brightest when they are in opposition both as cos E in-
creases and K decreases. In other words the Superior
planets appear more and more brilliant when they are
retrograding, being most brilliant at ‘C’. The spherical
radii of Jupiter Saturn and Mars being in decreasing
order, their brilliance will be in decreasing order so that
they will be rising at distances from the Sun which are in
increasing order. The inverse square law of courses works
here but combining the two factors (1) the spherical radius
of the planet and (2) the inverse square law, we may take
it that Bhāskara’s numbers cited above namely 28°, 17°
and 14° for Mars, Saturn and Jupiter respectively accord
with truth, for, these numbers are given by Bhāskara or
his authority Brahmagupta only after observing the planets
rising heliacally.

169
That the positions of the planets while setting or rising
heliacally will be situated symmetrically with respect to
the Sun, goes without saying.
But one thing. In the chapter called Udayāstamayā-
dhyāya, the degrees known as Kālāṁśas which are given
as the arcs between the planets and the Sun for heliacal
rising or setting, are different from the numbers given
above, for, the latter are the values of the Śīghra anomaly.
In the case of Mars when the Śīghra anomaly is 28°, the
Śīghraphala will be 11°, so that the apparent distance of
the planet will be 28° - 11° = 17° which are the Kālāṁśas
for Mars. Similarly in the case of Jupiter, when the
Śīghra anomaly is 14°, the Śīghraphala will be 3°, so that
the distance between the planet and the Sun as seen from
the earth will be 11°, which are given as Jupiter's Kālā-
ṁśas. Also in the case of Saturn, the Śīghraphala for 17°
of anomaly will be 2°, so that the Kālāṁśas would be 15°.
Verse 43. Mercury and Venus rise in the West by
50° and 24° of Śīghra anomaly respectively, and set in
the West by 155°, and 177° respectively. They rise in the
East by 205° and 183° of Śīghra anomaly and set there by
310° and 336° respectively.
Comm. Regarding the Inferior planets, they rise
heliacally in the East after Inferior conjunction and then
they are retrograde. They attain gradually the maximum
elongation in the East and after they revert to direct
motion, their elongation gradually decreases. They then
set in the East and heliacally rise thereafter in the West.
There again their elongation attains a maximum value;
then they begin to retrograde and gradually set in the
West only to rise in the East. This is all clear from the
heliocentric figure 11. When the Śīghra anomalies of
Mercury and Venus happen to be respectively 50° and 24°,
their Śīghraphalas would be 13° and 11° respectively, so
that they are themselves the Kālāṁśas, in as much as in
22

170
the case of the Inferior planets, the Sighraphala will be
itself their elongation eastern or western. Then they
rise in the West, being near Superior conjunction. When
again their Sighra anomalies equal respectively 155° and
177°, the same Sighraphalas will arise so that they set
heliacally in the West. Then as the Sighra anomalies
attain the symmetrical values on the other side ie. (360-
155) and (360-177) ie. 205° and 183° the Sighraphalas being
the same, they rise in the East. Again when they attain
the values (360-50) and (360-24) ie. 310° and 336°, they
set in the East on account of the same Sighraphalas or
Kālāmsas.
Verse 44. When the Sighra anomalies have parti-
cular values, to decide when the planets rise or set helia-
cally, we have to take the difference of those particular
values and the numbers given above for the respective
Sighra anomalies, convert them into minutes of arc and
divide the results by the daily motion in the Sighra ano-
malies in minutes of arc. Then we have the number of
days in which the rising or setting takes place thereafter.
Comm. Suppose it is required when Mercury rises in
the West. Suppose we want to compute this on a parti-
cular day when Mercury's Sighra anomaly is x°. Then
because we know that Mercury rises in the West when his
Sighra anomaly is 50, we have to calculate by how many
days the difference | x-50 | of the Sighra anomaly would
be covered. Let the daily motion of the Sighra anomaly
be y' per day, Then by rule of three (| x-50 | × 60) / y will
be the number of days before or after as the case may be
for Mercury to rise in the West.
Verse 45. To obtain the Mean planet knowing the
True.
Assume the True planet to be the Mean; compute the
Manda and Sighraphalas and applying them inversely,

171
we have an approximation of the Mean planets. Treating
these as the Mean planets, again obtaining the Manda and
Śīghraphalas and again applying them inversely and
repeating the process till constant values are obtained, we
have by this method of successive approximation the Mean
planets required.
Comm. The method of successive approximation is
clear.
Verse 46. To obtain the equinoctial shadow.
Convert the Ayanāṁśas into minutes of arc, and
divide by the mean daily motion of the Sun; then we
have the number of days before the Meṣa or Tulā Saṁ-
krānti day, or before Makara and Karkaṭaka Saṁkrānts
days, when the Sun will be in equinoxes or Solstices
respectively. The mid-day shadow of the Sun cast by the
gnomon on such an equinoctial day, will give us the
equinoctial shadow required.
Comm. The palabhā or equinoctial
shadow as it is called is the length of
the shadow cast by a gnomon taken
to be of 12 units in length, (measuring
the shadow also in the same units), at
noon of an equinoctial day. In other
words, if this shadow be of s units,
clearly s / 12 = tan ϕ (Vide fig. 18).
The Ayanāṁśas are the degrees of
the arc of the ecliptic in between the
Hindu Zero point of the Zodiac and the
first point of Aries which is now behind the former due to
the phenomenon known as the precession of the equinoxes.
They are called Ayanāṁśas because the solstices are also
behind the Makara and Karkaṭaka Saṁkrānti points of
the Hindu Zodiac or points which have Hindu longitudes
270° and 90° resply, by the same arc. The Hindu astro-

172
nomers came to know that the solstices are preceding by
observations made with the gnomonic shadow at mid-day
around the solstitial days. The day on which the maxi-
mum mid-day shadow is cast by the gnomon is the Winter
solstitial day whereas the day on which the mid-day
gnomonic shadow is maximum in the southern direction
(assuming the place to be of northern latitude and ϕ < ω)
or minimum in the northern direction (ϕ > ω) is the
summer solstitial day. Calculating the Sun's longitude
on that day at noon, we know how far the solstitial points
have preceded behind the points of the Hindu Zodiac which
have Hindu longitudes 90° and 270°. The word अयनांशाः
has therefore the meaning अयनविलोमगत्यंशाः where the
Samāsa may be viewed as a मध्यमपदलोपी समासः
Verses 47, 48. Calculating the five fundamental
H sines of a point of the Zodiac pertaining to a point of
the ecliptic.
The declination has to be computed from the sum of
the Hindu longitude of that point and the Ayanamsas.
Similarly if it be required to find the time before or after
the rise of a point of the ecliptic, we have to compute them
from the sum of the longitude of that point and the
Ayanamsas.
H sin 24° × H sin λ
─────────────────── = H sin δ I
R
√(R² — H sin² δ) = H cos δ = Dyujyā II
δ = H sin⁻¹ (H sin δ) III
s
── × H sin δ = Kujyā IV
12
Kujyā × R
───────── = Charajyā V
H cos δ
H sin⁻¹ (Charajyā) = Charam VI
Comm. (1) Let rA☉ be the ecliptic where r =
Vernal Equinox, A = first point of the Hindu Zodiac,

in between r and ☉,
not shown in the figure
☉ = the Sun. Let rA =
a° = Ayanamsas defined
before so that r☉ = (a + λ).
From the spherical triangle
r☉M, by Napier's rule,
sin δ = (sin λ + a) sin ω
which in Hindu trigono-
metry becomes H sin δ =
[H sin (λ + a) × sin ω] / R
In Hindu Astronomy the
obliquity of the ecliptic was
taken to be 24°. The value
of the obliquity is now 23°–27' approximately and it has
been know that this has been decreasing. At the time
when the Hindu Astronomers observed this, it should
have been greater than 23°–27' so that if it was taken to be
24°, their observations were not far from truth. This
means that the antiquity of Hindu Astronomy might be far
more than what the Moderns estimate it to be.
(2) How the formula I was derived in Hindu tri-
gonometry was as follows. Let ♋ be the summer solstice
♋C, ☉B, the perpendiculars dropped from ♋ and ☉ on
the line of intersection of the Equatorial and Ecliptic
planes namely rBC. Let perpendiculars be dropped from
♋ and ☉ on the plane of the Equator. Let them be ♋N,
☉D, Join NC and DB. Then ☉B̂D = ♋ĈN = the
dihedral angle between the two planes = ω; ♋C = R
since in Hindu trigonometry H sin 90° = R. Also ♋N =
H sin ♋E (in Hindu trigonometry) = H sin ω; ☉D =
H sin ☉M = H sin δ; ☉B = H sin r☉ = H sin (λ + a)
where λ is the Hindu longitude of the Sun and a =
Āyanamśas. It will be seen that ☉D is a segment of the

line of intersection of the planes PCM a plane perpendi-
cular to the plane of the Equator, and ☉BD a plane per-
pendicular to the Ecliptic plane. Since ⚍C 11 ☉ B and
⚍N 11 ☉ D and BD☉ = CN⚍ = 90°, so the two tri-
angles are congruent
∴ ☉B / ⚍C = ☉D / ⚍N ie. H sin (λ + a) = (R × H sin δ) / (H sin ω)
∴ H sin δ = [H sin (λ + a) H sin ω] / R
NC = ⚍L = √(⚍C² — ⚍N²) = √(R² — H sin² δ) =
H cos δ. ⚍L is called Dyujyā as explained below in
note (3).
Another way of looking at the similarity of the tri-
angles C⚍N and B☉D is from the fact that they are
formed by the intersection of parallel planes, both of which
are perpendicular to the plane of the Equator. This idea will
be elaborated when we explain fig. 21, wherein the so-called
latitudinal triangles will be shown to be formed by the
intersection of the plane of the Equator and planes of
diurnal circles with the planes of the horizon and the
prime vertical. In fact, the plane of a great circle and
the parallel planes of the corresponding small circles form
the same dihedral angle with the planes of the celestial
sphere namely the planes of the meridian, horizon and
prime-vertical so that right-angled triangles formed by
their intersection will be all similar.
Formula II is derived from the formula H sin² δ +
H cos² δ = R² derived from fig. 6 from which formula,
formula III follows.

176
(3) Why H cos δ is called Dyujyā in formula II
is clear from fig. 20 wherein p is the celestial pole, (C) is the
Fig. 20
celestial Equator and (C) is the diurnal circle of a celestial
body say the ☉ ie. the Sun. Let ☉N be the perpendicular
dropped from ☉ on the plane of the Equator so that
☉N = H sin ☉A = H sin δ CN = H cos δ = C☉ =
radius of the diurnal circle called Dyujyā (Dyu = day)
द्युवृत्तत्रिज्या = द्युज्या (Madhyamapadalōpi Samāsa) formulae
IV, V and VI will be dealt with in the next chapter
Triprasnādhyāya more elaborately but one has to under-
stand what Charajyā is to follow the subsequent verses of
this chapter so that we shall give its location and defini-
tion in the light of fig. 21. Let fig. 21 represent the
celestial sphere in which SEN = Horizon, Z = Zenith,
N = Nadir Ez = prime vertical EQR = celestial equator.

176
Fig. 21
PP′ = Polar axis, SBS′ = the diurnal circle of a celestial
body S; EW = the East-West line, SS′ = Udayāstasūtra
or the join of the rising and setting points S, S′. This
SS′ is evidently a diameter of the diurnal circle which is
bisected by the plane of the horizon; PEP′ is called the
unmandala or the Equatorial horizon. PSA is the decli-
nation circle of S cutting the Equator in A. EA is called
the charam whose H sine is called Charajyā. The H sine
of SB in the diurnal circle is called Kujyā so that as
corresponding lines in the diurnal circle and the Equator
stand in the ratio H cos δ : R,
Kujyā / Charajyā = (H cos δ) / R I