ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

124 INDIAN LUNI-SOLAR ASTRONOMY
where D' means the Moon as corrected by the 1st and the 2nd
equations. Here the constant is wrong by −1' 18".
(4) His fourth equation or the annual equation
= ± 1/10 of the Sun's equation of apsis,¹
= ± 1/10 × (12 × 3438) / 360 sin (Sun's distance from the
apogee).
= ±11'27". 6 sin (Sun's distance from the apogee).
The modern value of the constant is 11' 10". Tycho found
it to be 4' 30". Horrocks' (1639) co-efficient was 11' 51".
As Candraśekhara was aware of Bhāskara's Bījopanaya, as
also of the work of Śrīpati, his merit here lies in the discovery
of the annual equation, and correction to the constant of
variation.
Thus we have seen that so far as the luni-solar astronomy
is concerned Indian astronomy is independent of Greek astronomy
in respect of astronomical constants, that Indian astronomy is
generally more accurate than Greek astronomy and that Indian
astronomers were not mere "calculators"². There were observers
who verified and corrected the old astronomical constants as
they came down from Āryabhaṭa and Brahmagupta, who also
found independently all the principal equations of the Moon.

1. Siddhānta darpaṇa VI. 13.
2. G.R. Kaye Hindu Astronomy, p. 60
   — : o : —
   Reference
   P.C. Sengupta : The Khaṇḍakhādyaka, 1934
   Karl Manitius's edition : Syntaxis.
   Godfray : Lunar Theory.

CHAPTER VII
Greek and Hindu
Methods in Spherical Astronomy
Here we shall reproduce from Sengupta's paper a compa-
rative account of the Greek and ancient Indian methods in
Spherical Astronomy and to bring out the independence of the
Indian Astronomers on this subject. The views on this subject
would necessarily differ from those of many European scholars
such as Colebrooke and Bentley (early 19th century) to Kaye
(early 20th century). Kaye wrote as follows in the Journal of
Asiatic Society of Bengal 1919, No. 3.
The methods by which (the rules) were obtained are
buried in obscurity. Braunmühl¹ has stated "that the
Indians were the first to utilise the method of projec-
tion in the Analemma of Ptolemy." It is intended to
present the Hindu methods as clearly as possible and
to show that Braunmühl has not done sufficient justice
to the Indian astronomers.
As to Kaye, we shall show that his remark quoted above
is due to the fact that he had to rely mostly on the English tran-
slation of the Sūryasiddhānta of Burgess, and perhaps he had no
access to the works of Bhāskara II (1150 A.D.), who was the first
to explain the ancient Indian methods clearly.
Greek and Hindu Methods in Spherical Astronomy
Of the Greek methods in Spherical Astronomy, the
history begins with elementary principles only from Euclid (300
B.C.). Even in Theodosius' Sphaerica² (about 153 B. C.) "there
is nothing that can be called trigonometrical." Heath again says,

1. Heath, Greek Mathematics, Vol. II, p. 291. Braunmuhl, Geschichte der
   Trigonometrie, pp. 38-42.
2. Heath, Greek Mathematics, Vol. II, p. 250.

126 GREEK AND HINDU SPHERICAL ASTRONOMY
"the early spheric did not deal with the geometry of the sphere as
such, still less did it contain anything of the nature of the spheri-
cal trigonometry. (This deficiency was afterwards made good by
Menelaus's Sphaerica).¹ Hence the Greek spherical trigonometry
began with Menelaus (90 A.D.). His theorem in geometry is
well-known—"If the sides of a plane triangle be cut by a trans-
versal into six segments, the continued product of any three
alternate segments, is equal to the continued product of the
remaining three." From this proposition he deduced the so-
called "regula sex quantitatum" or the theorem, if the sides of a
spherical triangle be cut by an arc of a great circle into six
segments, the continued product of the chords of the doubles of
any three alternate segments is equal to the continued product
of the chords of doubles of the remaining three segments." In
plane geometry if the sides BC, CA, AB of a triangle be cut by
any transversal at L, M, N, respectively,
then we have
(BL / LC) · (CM / MA) · (AN / NB) = 1.
In spherics the theorem is :
(Chord 2 BL / Chord 2 LC) · (Chord 2 CM / Chord 2 MA) ·
(Chord 2 AN / Chord 2 NB) = 1
Both these theorems are proved in
Ptolemy's Syntaxis ( Karl Manitius's edi- Fig. 5
tion, Vol. I, pp. 45-51).
If R be the radius of the sphere on which the spherical
triangle ABC is constructed, then the chord of the arc 2 BL =
2 R sin BL. Hence Menelaus's theorem in spherics may be expres-
sed as follows :
(Sin BL / Sin LC) · (Sin CM / Sin MA) · (Sin AN / Sin NB) = 1
This theorem is true for any spherical triangle.
If ∠B = AN = AM = 90° and L the pole of AB, then LMN
is a secondary to the arc AB. There are four arcs of great
circles; taking any three as forming a spherical triangle and the
fourth as the transversal we readily get for the right-angled

1. A.A. Bjornbo, "Studien über Menelaos' Sphärik" in Abhandlun-
   gen Zur Geschichte der Mathematischen Wissenschaften for 1902, pp. 89 et
   seq.; also Heath, Greek Mathematics, vol. II p. 261-73.

GREEK METHODS IN SPHERICAL ASTRONOMY 127
triangle ABC, the relations :--
(i) sin a=sin b sin A
(ii) sin c=tan a cot A
(iii) cos b=cos a cos c
(iv) tan c=tan b cos A
The above are some of the Napier's rules for a right-angled
spherical triangle, deducible from Menelaus's theorem¹. They
are generally sufficient in the case of such triangles. In any
spherical triangle, however, this theorem of Menelaus does not
in any single step lead to any of the equivalents of the time-
altitude or altazimuth equations in spherical astronomy. The
ancient Indian methods, though none of them are so highly
finished as Menelaus's theorem, yet are not less powerful in
tackling the problems that arise in astronomy in connection
with the apparent diurnal motion of the heavens. The Greek
or Ptolemaic method presents no further points of interest except
in its application. We now proceed to illustrate the ancient
Indian methods and shall refer to the Ptolemaic method as
occasion arises.
Ancient Indian Methods
in Spherical Astronomy
In the Indian methods there is no general rule to follow.
It is by properties of similar right-angled triangles that a fairly
complete set of accurate formulae are obtained. These right-
angled plane triangles are classified under the names,—'Krānti-
kṣetras' (triangles of declination) and 'Akṣa-kṣetras' (triangles of
latitude). We consider the following problems :—
Problem :— To find the time of rising on the equator of a
length l, of arc of the ecliptic measured from the
first point of Aries.
Let ω be the obliquity of the ecliptic and R. A. the right
ascension corresponding to the longitude l, and δ the correspond-
ing declination. The Indian form of the equation is :

1. Three more can be deduced similarly, namely,
   (v) sin c=sin b sin C
   (vi) sin a=tan c cot C
   (vii) tan a=cos C tan b.

128 GREEK AND HINDU METHODS
*R sin R.A. = (R sin l × R sin ω) / (R cos δ) , where R is the radius of
the sphere.
Note:— If R be the radius of the circle of reference, the
Indian trigonometrical functions for the arc θ, are
(1) the 'sine,' (2) the 'cosine' and (3) the versed
sine. They are respectively equal to R sin θ,
R cos θ and R vers θ.
In the adjoining figure, O is
the centre of the armillary
sphere, YQ, YC are quadrants of
the equator and the ecliptic,
respectively. P is the celestial
pole, PCQ the summer solsti-
tial colure. Join OY, CQ, OP
and OC.
Let YS be=l, YM=R.A.,
CQ=∠SYM=ω, SM=δ.
Join OS, OM. PSM is the
secondary to the equator.
[Fig. 6]
From C draw CK perpendicular to OQ. From S draw Sm
and Sn perpendicular to OM and OY, respectively. Join MN
and from M draw MN perpendicular to OY.
Then the triangles Smn and CKO are similar. They are
called 'Krānti-kṣetras'¹ or declination triangles,—similar right-
angled triangles having one acute angle=ω.
∴ Sm : Sn = CK : OC
or R sin δ : R sin l = R sin ω : R
R sin δ = (R sin l × R sin ω) / R ......(I)

The Āryabhaṭīya, Gola, 25. Varāhamihira, in the Pañcasiddhāntikā
(IV. 92) states it in the form 2R √((R² Sin ²l)—R² Sin ²δ) / (2R cos δ)
= R Sin R.A., which is evident from figure. Brahmagupta's
equation is identical with that of Āryabhaṭa, (BrSpSi. III. 15, Sūrya-siddhānta
III. 40-41.
Also Bhāskara II, Grahagaṇita cap. VIII, stanzas 54-55, is in agreement
with Varāhamihira's forms.

TIME OF RISING 129
Greek Method
In the same figure¹ let PSC be the triangle and γMQ be the transversal. Then Menelaus's theorem gives
(sin PM / sin MS) × (sin Sγ / sin γC) × (sin CQ / sin QP) = 1
or (1 / sin δ) × (sin l / 1) + (sin ω / 1) = 1
or sin δ = sin l × sin ω.
Indian Method
Again by the Indian method from the same two similar triangles we get
mn : nS = OK : OC
or, mn : R sin l = R cos ω : R
∴ mn = (R sin l × R cos ω) / R
Again MN : mn = OM : Om
i.e., R sin R. A. : mn = R : R cos δ
∴ R sin R.A. = (R sin l × R cos ω) / (R cos δ)
Greek Method
Take PQM for the triangle and γSC for the transversal.
Then,
(sin PC / sin CQ) × (sin Qγ / sin γM) × (sin MS / sin SP) = 1
or (cos ω / sin ω) × (1 / sin R.A) × (sin δ / cos δ) = 1
or sin R.A. = tan δ cot ω,
The Indian form of the equation is different from that of Ptolemy's. It is also better for the purpose of calculation.
Note :—From the same two similar triangles we have
On : ON = R cos δ ; R
∴ On : R cos l = (R cos R.A. × R cos δ) / R ......(3)
Again, tan R.A. = mn / on
= (R sin l × R cos ω) / (R × R cos l) ......(4)
Again, mn : Sm = OK : KC

1. Manitius' Edition of Syntaxis, I, 51-53.

130 GREEK AND INDIAN METHODS
or mn = (R sinδ × R cosω) / (R sin ω)
∴ R sin R. A. = (MN / mn) × mn = (R / R cosδ) × ((R sinδ × R cosω) / (R sin ω)) (5)
Problem II :—
Sidereal Time-intervals
Indian Method
The problem discussed above provides the method of find-
ing the sidereal time-intervals in which the signs of the zodiac
rise on the equator. To find the corresponding times at any
latitude ϕ, it is necessary to calculate and apply what is the
ascensional difference due to the elevation of the celestial pole.
This ascensional difference is called 'carakāla' or the variation in
the length of half the day. The 'sine' of this 'carakāla' is called
'carajyā.' If ch denotes this 'carakāla,'
then.¹ R. Sin ch = (R sinϕ × R sin δ × R) / (R cos ϕ × R cos δ)
Just as in the solution of the previous problem, the dec-
linational triangles or 'Krānti Kṣetras' were constructed and
used, so in the solution of this and other problems another set
of similar triangles were conceived and constructed and were
given the name 'Akṣa kṣetras.'²
Let NPZH be the meri-
dian (Fig. 7), NOH the north-
south line passing through
the observer O,P the celestial
pole, OQ the trace of the
equator on the meridian
plane, Z the zenith. Join OZ.
From Q draw QM perpendi-
cular to OZ. Then the
triangle QOM is an 'Akṣa-
kṣetra' or a latitudinal right-
angled triangle, as ∠QOM =
ϕ, the latitude of the station.
Another 'Akṣakṣetra' is thus
conceived, in the same figure, let P, P' be the north and
south celestial poles, N, the north point, AB A'B' the diurnal
Fig 7

1. Āryabhaṭīya, Gola, 26; Pañca-siddhāntikā, IV, 34; Brāhmasphuṭasiddhānta.
   II, 57-58; Sūrya-siddhānta, II, 91; Grahagaṇita, VIII, 48-49.
2. Bhāskara, Golādhyāya ( Wilkinson and Bāpudeva Śāstri's tr. ) PP. 173-76;
   also, Bhāskara, Grahagaṇita, Ch. IX. 13-17.

SIDEREAL TIME-INTERVALS 131
circle of a heavenly body with declination δ, NEHW the
horizon, PEP′ W the six O′ clock circle. Here AA′ the line of
intersection of the diurnal circle with the horizon is called the
“udayāsta-sūtra”¹ (or the thread joining the rising and setting
points). SS′ the line of intersection of the diurnal circle and
the six o′ clock circle, is the horizontal diameter of the diurnal
circle. From S draw SK and SL perpendiculars respectively to
AA′ and EW. Join KL.
Now since PN=ϕ, the latitude of the station, in the small
right-angled triangle KLS, the ∠ KLS is also=ϕ.
∴ SK : SL=QM : MO
or SK = (SL × QM) / MO = (R sin δ × R sin ϕ) / (R cos ϕ)
Now SK² is a “sine” in the small circle AB A′B′ of which
the radius is R cos δ; this “sine” reduced to the equator (radius R)
is the ‘sine’ of cara.
∴ R sin ch = R sin EPA
= (R sin δ × R sin ϕ × R) / (R cos ϕ × cos δ)
Greek Method
Let³ the arc PA be produced to meet the equator at C.
Take PCQ′ for the triangle and EAN for the transversal. Then
we get,
(sin PA / sin AC) × (sin CE / sin EQ′) × (sin Q′N / sin NP) = 1
or (cos δ / sin δ) × (sin CE / 1) × (cos ϕ / sin ϕ) = 1
∴ sin CE = sin ch = (sin ϕ × sin δ) / (cos ϕ × cos δ)
Note—The perpendicular distance between AA′ and Ew is
called the ‘sine’ of the amplitude or the ‘Agrā’ which is thus
calculated :—
KL : LS = QO : OM
∴ ⁴R sin amplitude = ‘Agrā’ = KL
= (LS × QO) / OM = (R sin δ × R) / (R cos ϕ)
It is now evident that the ancient Indian method is different
────────────────────────────────────────

1. Bhāskara, Gola, VII, 39.
2. This is called by the name ‘kujyā’ or ‘kṣitijyā’. i, e.. earth-sine. Āryabhaṭa,
   Gola, 26, Brahmagupta, II, 57, Sūrya-siddhānta. II, 61 etc.
3. Manitius, ibid, p. 84.
4. Āryabhaṭa, Gola. 30, etc.

132 GREEK AND INDIAN METHODS
from the Greek method in this case also. As the triangle KLS
is difficult to show in the diagram, it is shown in its projection on
the meridian plane in Burgess's translation of the "Sūrya-
siddhānta,"(page 232)and also in Wilkinson and Bāpūdeva Śāstrī's
translation of the 'Siddhānta Śiromaṇi,' p. 175. This has led
Braunmühl to assume that the Indian method of arriving at the
equation of ascensional difference and some other equations of
spherical astronomy has its origin in the Analemma of Ptolemy.
A careful study, however, does not justify the identification of
Indian methods with the graphic method of the Analemma,
which is deduced from the projections of the position of a heav-
enly body on the meridian prime vertical and the horizon. It is
being presently shown that what was done out of difficulty in
drawing the figures properly has been taken by Braunmühl as a
Greek connection.
Problem III¹ :—
To find the “Time-altitude” Equation
If from any point S on the diurnal circle a perpendicular
be drawn to the Udyāsta-Sūtra spoken of before, this perpendi-
cular is called the cheda or ‘iṣṭahṛti.’ The perpendicular from S
on the horizon is called ‘Śaṅku’² the sine of the altitude. The
line joining the foot of the ‘Śaṅku’ and that of the perpendicular
on the ‘Udayāsta-Sūtra’ goes by the name of ‘Śaṅkutala’ and this
Śaṅkutala lies to the south of the ‘Udayāsta-Sūtra’ during the day.
In this figure (Fig 8) if AA' be the ‘Udayāsta-Sūtra’ or the
intersection of the diurnal circle and the horizon, and S a point
on the diurnal circle denoting a position of the Sun, SK, SL
perpendiculars on AA' and the horizon respectively; SL is called
the ‘Śaṅku,’ SK the ‘cheda’ and LK, the ‘Śaṅkutala’. In this
triangle KSL, the angle KSL was recognised to be the latitude of
the station.
Thus the triangle SKL is not taken in its projection on the
meridian plane. The side SK is taken 'as formed of two parts.

1. Āryabhaṭa could not arrive at the true equation. Cf. Gola 28. The correct
   rules occur in Pañcasiddhāntikā, IV, 42, 44; Brahmasphuṭasiddhānta, III,
   36-38, 26-40; Sūryasiddhānta, III, 34-35.
2. Bhāskara says : ग्रहस्थानाल्लम्बः शंकुः । तस्यतलमुदयास्तसूत्राद्दक्षिणतो भवति ॥
   “Gola, VIII-39-41, Āryabhaṭa uses the term शङ्क्वग्रम्” Gola, 29.

TIME-ALTITUDE EQUATION 133
Let CC' be the line of inter-
section of the diurnal circle
and the 'six o'clock 'circle
EPW. Let SK cut CC' in
M. Then.
SK = SM + MK
Here SM, the 'sine' in
the diurnal circle of the
complement of the hour angle
is given a distinct name
'Kāla'¹ and MK as explained
before is known by the name
Fig. 8
'Kujyā.' This 'Kāla' is constructed from the point S in the diurnal
circle. Thus the triangles like SKL were not taken in their
projections on the meridian plane as Braunmühl would suggest.
From the triangle KSK, we get,
'Cheda' : 'Śaṅku' = R : R cos ϕ where ϕ is the latitude of the
observer;
'Śaṅku' is here = R cos Z, Z being fhe Sun's zenith distance.
∴ 'cheda' = (R cos Z × R) / (R cos ϕ)
Now 'Cheda' = radius of the diurnal circle + Kujyā - versed
sine of the hour-angle in the diurnal circle O' B + O' V - BR,
= R cos δ + (R sin δ × R sin ϕ) / (R cos ϕ) - (R vers H × R cos δ) / R
As in the previous problem, Kujyā = SK = (R sin δ × R sin ϕ) / (R cos ϕ)
or (R cos Z × R) / (R cos ϕ) = (R cos δ / R) { R + (R sin δ × R sin ϕ) / (R cos ϕ)
× R / (R cos δ) - R vers H }
The above equation simplified becomes
cos Z = sinδ sin ϕ + cos δ cos ϕ cos H.
In this connection we consider the altazimuth equation by
the Indian method.

1. Bhāskara's Grahagaṇita, VIII, 55.
   O' is the middle point of CC' or it is the centre of the diurnal circle
   ABB'.

134 GREEK AND INDIAN METHODS
¹Problem IV :—
The Altazimuth Equation
Indiad Method
Let α denote the azimuth of the Sun from the south. In
the same triangle SKL in the same figure, we have,
LK : SL = R sin ϕ : R cos ϕ
or, ‘Śaṅkutala’ : ‘Śaṅku’ = R sin ϕ : R cos ϕ
∴ ‘Śaṅkutala’ = (R cos Z × R sin ϕ) / (R cos ϕ)
Now ‘Śaṅkutala’ is made up of two parts, namely, ‘Bāhu’
and ‘Agrā’, of which the former is the distance of L from the
observer’s East-West line; the ‘Agrā’ has been already found.
Here ‘Bāhu’ = (R sin Z × R cos α) / R and ‘Agrā’ = (R sin δ × R) / (R cos ϕ)
∴ ‘Śaṅkutala’ = ‘Bāhu’ + ‘Agrā’
or (R cos Z × R sin ϕ) / (R cos ϕ) = (R sin Z × R cos α) / R + (R sin δ × R) / (R cos ϕ)
or R sinδ = (R cos ϕ / R) ((R cos Z × R sin ϕ / R cos ϕ) - (R sin Z × R cos α / ϕ))
which is easily seen to be equivalent to
sin δ = cos Z sin ϕ - sin Z cos ϕ, cos α
Greek Method
Ptolemy² has also a method of finding the Sun’s altitude at
any hour of the day. His method is as follows :—
(i) He would find by
means of his tables for the
times of risings of the signs of
the zodiac, the orient ecliptic
point. (ii) He would then
find the culminating point of
the ecliptic. (iii) He would
finally apply Menelaus’s
theorem in spherics thus :—
Fig. 9
Let ASC be any position
of the ecliptic, (Fig. 9) NZC the

1. The equivalent of this, in a particular case, is first found in Brāhmasphuṭasiddhānta, Ch. III, 54-56 Cf. Sūryasiddhānta, III, 28-31, also Bhāskara Grahaganita, IX, 50-52.
2. Manitius, ibid, pp. 118, 19.

PTOLEMY'S ANALEMMA 135
meridian, NAMH the horizon, Z, the zenith and S the Sun.
Here the celestial longitudes of C, S and A are taken to be
known; hence ZC and CH are also known.
Now take ZCS for the triangle and HMA to be the trans-
versal ; we then have by Menelaus's theorem.
(sin ZH / sin HC) × (sin CA / sin AS) × (sin SM / sin MZ) = 1
or sin SM = (cos CZ × sin AS) / sin CA
It is thus clear that Ptolemy had no direct method for
connecting the Sun's altitude and the hour-angle. This method
is workable for the problem “given time, find the altitude” but
is not workable in the converse problem ; besides, the calcula-
tion of the longitudes of A and C is very cumbrous.
Again, when EA has been found out, taking ZHM for the
triangle and CSA for the transversal, we get,
(sin HA / sin Am) × (sin MS / sin SZ) × (sin ZC / sin CH) = 1, whence and thence HM,
the azimuth can be found. The method is here also cumbrous,
there being no direct connection between altitude and azimuth ;
besides the time-element is not avoided.
The Analemma of Ptolemy
and the Indian Method.
When the Sun's declination is zero and his hour-angle, is
H, Zeuthen¹ following the method of the ‘Analemma’ of
Ptolemy, as explained by Braunmühl² has deduced the following
equations :
(1) cos Z = cos H. cos ϕ
(2) tan α = tan H / sin ϕ
To these two, Heath following Braunmühl, adds
(3) ³tanZQ = tan H / cos ϕ
─────────────────────────────────────────────────────────────

1. Heath, Greek Mathematics, Vol. II, pp. 290-91.
   Zeauthen, Bibliotheca Mathematica, 13, 1900, pp. 23-27.
2. Braunmuhl , ibid, pp. 12-13.
3. The Indian form of this equatiom is R Sin ZQ = (R Sin H × R) / [√(R² - R²cos²H × R² Sin²ϕ) / R]
   Bhāskara's, Golādhyāya, Com. on VIII, 67.

136 GREEK AND INDIAN METHODS
where Z is the zenith and Q is the point of intersection of the
prime vertical and its secondary passing through the Sun and
the north-south points.
Zeuthen¹ points out that later in the same treatise Ptolemy
finds the arc 2β described above the horizon by a star of given
declination δ' by a procedure equivalent to the formula.
(4) cos β=tan δ' tan ϕ.
With regard to the 'Analemma' of Ptolemy. it may be
noted, as Heath² says, that "the procedure amounts to a
method of graphically constructing the arcs required as parts of
an auxiliary circle in one plane." Many things may be, in
practice, done graphically far more easily than by the theoreti-
cal method. Besides, no theoretical calculations occur in the
'Analemma'. Zeuthen², following the method of this work, has
deduced in the general case, the two equations.
(5) cos Z=(cos δ, cos H+sin δ. tan ϕ) cos ϕ.
cos δ.sin H
(6) tan α=————————————————————————————————————————
sin δ
————— +(cos δ.cos H+sin δ.tan ϕ) sin ϕ
cos δ
These equations are suggested to a modern reader from a
study of the figures in the 'Analemma.' But neither in this work
nor in the 'Syntaxis' are they to be found. With regard to the
first four formulae, it is possible that they were recognised by
Ptolemy. With regard to the last two, Zeuthen³ remarks "mais
le texte nen contient rien,' and they were certainly not recog-
nised by Ptolemy.
Besides the tangent function is wholly absent in Greek
trigonometry. They are also different in form from those arrived
at by the Indian method as explained before. Thus, it is clear
that the Indian methods are in no way connected with the
method of the 'Analemma.'
Even taking for granted that the Indians followed a method
of projection much allied to the method of the Analemma' there
is no adequate reason for assuming that their method is derived
from any Greek source. Analogy and precedence do not neces-
sarily constitute originality—there is still the chance of a remoter
origin from which both the systems drew their inspiration. The
method of the 'Analemma,' as has been already stated, presents a
——————————————————————————————————————
1, 2, Bjornbo, loc. cit. p. 86.
3. Zeuthen, loc. cit. p. 27.

ANGLE BETWEEN ECLIPTIC AND MERIDIAN 137
graphical method for constructing the Sun's altitude and azimuth
from the hour angle when the Sun's declination is zero but such
a graphical method is generally complex as compared with the
elegant Indian method. An astronomer who constructs and uses
an armillary sphere to arrive at his equations in spherical astro-
nomy and who has not a well-developed spherical astronomy at
his command must have to draw perpendiculars from the positions
of the heavenly body, not only on the meridian plane, the hori-
zon or on the prime vertical, as the occasion arises, but also on
the line of intersection of the diurnal circle with the horizon.
Hence Braunmühl's statement that the Indian methods of spheri-
cal astronomy have their origin in the 'Analemma's, in spite of
his admitting that Indians were first to utilise its methods, is rather
far-fetched and tends to take away the honour from the great
Indian astronomers, who devised the beautiful methods. The
'Analemma' as it now exists is a Latin translation from an Arabic
version of the original Greek¹. We may reasonably doubt that the
Arabic version was greatly influenced by the ancient Indian system.
We now pass on to the consideration of other allied or
similar problems in the two systems of astronomy.
Problem V—
To find the Angle between the Ecliptic and the Meridian
Indian Method²
Let ♈SA be thee cliptic, ♈CE
the equator, E the east-point of
the horizon (Fig 10). Cut off
SH=90° and draw the great
circle HEAP' cutting the meri-
dian P'SCH at the points P' and
H. The aim is to find AP' but
it is enough to find EA since AP'
is the complement of EA'.
Both Āryabhaṭa and
Brahmagupta were unable to
find EA correctly. Let P be the
celestial pole and let PAE' be

1. On the influence of the ancient Indians on Arab mathematics and astronomy; see Alberuni's India, translated by Dr. E. Sachau, Vol. II, p. 304.
2. Āryabhaṭa, Gola, 45; BrSpSi, IV. 17; Sūrya-siddhānta, IV. 25; Bhāskara's Golādhyāya, VIII, 21-74, first example in his own commentary.

138 GREEK AND INDIAN METHODS
the secondary to the equator cutting it at E'. Both the above
astronomers were content with the idea that AE = AE', or that
AE = the declination of the point A of the ecliptic which is 90°
ahead of S in the above figure. This idea continued till the
time of Bhāskara II (1150 A. D.) who found out the correct
equation.
He recognised that CS, the declination of S = PP'; P'EH is
then the horizon of the station whose north geographical latitude
is CS. Also, the ‘sine’ of EA is the ‘Agrā’ or the sine of the
amplitude of the point A for the latitude CS.
∴ R sin EA = (R sin AE' × R) / (R cos CS) = [R sin (90° + γS) × R sin ω] / R
× R / (R cos CS)
or R sin EA = [R sin (90° + l) × R sin ω] / (R cos δ)
where l stands for γS and δ for CS.
Greek Method :
We give below the Ptolemy's method in a slightly modified
form¹. Let SHA be the triangle and γCE be the transversal ;
then we have,
(sin SC / sin CH) × (sin HE / sin EA) × (sin Aγ / sin γS) = 1
or (sin δ / cos δ) × (sin 90° / sin EA) × [sin (90° + l) / sin l] = 1
∴ sin EA = [sin δ × sin (90° + l)] / (cos δ × sin l) ,
which is readily transformed into Bhāskara's equation. The
originality of Bhāskara would be readily admitted.
Problem VI--
To find the Angle between the
Ecliptic and the Horrizon
Indian Method :
(A) Āryabhaṭa's method. It consists of the following²
steps :--
(1) Determination of the orient point of ecliptic.
(2) Finding the sine of its amplitude.

1. Manitius, ibid, Book I, pp. 104-06.
2. Āryabhaṭa, Gola, 33 : Sūryasiddhānta, V. 5-6.

ANGLE BETWEEN ECLIPTIC AND HORIZON 139
(3) Determination of the culminating point of the ecliptic
from the hour-angle of the Sun.
(4) Finding the declination of the culminating point of the
ecliptic.
Having obtained the above elements, his rule can be follow-
ed thus :
In this Fig. 11 NZH is the meridian, HMEAN the horizon,
CN'A the ecliptic. If N' be the nonagesimal or the highest point
of the ecliptic, the altitude of N' is the inclination of the ecliptic
to the horizon.
Let ZN'M be the vertical through N', meeting the horizon at M.
When the time is given, the longitudes of A and C can be
found out, from which CZ the zenith distance of C and EA the
amplitude of the orient ecliptic point can be determined.
Fig. 11
Here HM=EA.
According to Āryabhaṭa,
R sin CN' = (R sin CZ × R sin HM) / R
and R sin ZN' = √((R sin CZ)² - (R sin CN')²)
This is only an approximate rule. As expressed here,
R sin ZN' = (R sin CZ × R cos HM) / R approximately.
= (¹R sin CZ × R cos HM × R) / (R × R cos CN') accurately.
= (R sin CZ × R cos HM) / (R cos CN') .
(B) The method of Brahmagupta² :
Brahmagupta would also first determine the orient ecliptic

1. This correction was perhaps first noticed by Raṅganātha (1603 A. D.)
   in his commentary in the Sūryasiddhānta.
2. BrSpSi. V 3.

140 GREEK AND INDIAN METHODS
point A. Then he subtracts 90° from the longitude of A. Thus
having the longitude of N', he next finds the part of the day
elapsed of N' ; from which by the time-altitude equation discussed
above, he finds ZN'. This is of course more accurate than that of
Āryabhaṭa. Bhāskara¹ here follows Brahmagupta.
Greek Method :
Let the ecliptic CN'A cut the lower half of the meridian at
F. Ptolemy takes AK along the ecliptic=90° and AR along the
horizon=90°; then the great circle passing through R and K passes
through the nadir Z'. Now take Z'FK for the triangle and ANR
for the transversal, then by Menelaus's theorem.²
(sin FN / sin NZ') × (sin Z'R / sin RK) × (sin KA / sin AF) = 1
∴ sin RK = (sin FN / sin AF) = (cos FZ' / sin AC) = (cos CZ / sin AC) = (sin CH / sin AC)
or sin MN' = sin CH / sin AC.
Here Ptolemy's equation is simpler than that of Āryabhaṭa;
hence they must be independent of each other.
[Figure 12] Problem VII:—
To find the Angle made by the
Vertical through any Point of the
Ecliptic with the Latter
This problem is considered
by Ptolemy but it is not consider-
ed separately in Indian Astronomy,
but from the rule for parallax in
longitude, the rule for its calcula-
Fig. 12 tion can be deduced.
Indian Method :
In Fig. 12 S represents the true position of the
Sun and S' the Sun's position as depressed by parallax. N'SA is
the ecliptic. If from S', S'Q be drawn perpendicular to the ecli-
ptic, then, if P is the horizontal parallax,

1. Grahaganita; XII, 3-4.
2. Manitius, ibid, pp. 110-111.

ANGLE MADE BY THE VERTICAL 18
SQ = SS' × (R cos S'SQ / R) = (P × R sin ZS / R) (R cos S'SQ / R)
¹= (P / R) √[(R sin ZS)² – (R sin ZN')²]
²= (P / R²) × R sin N'S × R cos ZN', where N' is the nonagesimal
Thus R cos S'SQ is seen to be
= (R sin N'S × R cos ZN') / (R sin ZS)
The Indian method is fully described by Bhāskara in his
'Golādhyāya. VIII, 12-25. The truth of the Indian rule for R cos
S'SQ is easily seen from the spherical triangle ASN, where A is
the pole of the ecliptic.
Greek Method :
³Ptolemy takes SK and SL 90° each, along the vertical
circle ZSEK and the ecliptic N'SA. The great circle through K
and L cuts the horizon at R which is the pole of the vertical
circle. He takes SKL for the triangle and EAR for the trans-
versal, then
(sin SE / sin EK) × (sin KR / sin LR) × (sin LA / sin AS) = 1
or sin LR = (cos ZS × cos AS) / (sin ZS × sin AS)
or cos S'SQ = cot ZS × cot AS = tan SE × cot AS.
The Indian and the Greek rules are altogether different
both in form and method. There can, therefore, be no question
of any connection between them.
Problem VIII :—
To convert the Celestial Longitude of a Heavenly
Body into its Polar Longitude
If σ be the position of a (Fig.13), γK and σK are the celestial
longitude and the celestial latitude, respectively : γM and σM
are the polar longitude and polar latitude : γN and σV are the
right ascension and declination of the star.
Indian Method :
All Indian astronomers attempt at finding MK which, sub-

1. Āryabhaṭa, Gola, 34; Pañcasiddhāntikā, IX, 22 BrSpSi, XI, 23.
2. BrSpSi. V, 4-5 ; Sūryasiddhānta, V, 7-8 Bhāskara, Grahagaṇita, XII, 4.
3. Manitius, ibid, p. 119.

140 GREEK AND INDIAN METHODS
point A. Then he subtracts 90° from the longitude of A. Thus
having the longitude of N', he next finds the part of the day
elapsed of N' ; from which by the time-altitude equation discussed
above, he finds ZN'. This is of course more accurate than that of
Āryabhaṭa. Bhāskara¹ here follows Brahmagupta.
Greek Method :
Let the ecliptic CN'A cut the lower half of the meridian at
F. Ptolemy takes AK along the ecliptic = 90° and AR along the
horizon = 90°; then the great circle passing through R and K passes
through the nadir Z'. Now take Z'FK for the triangle and ANR
for the transversal, then by Menelaus's theorem.²
(sin FN / sin NZ') × (sin Z'R / sin RK) × (sin KA / sin AF) = 1
∴ sin RK = (sin FN / sin AF) = (cos FZ' / sin AC) = (cos CZ / sin AC) = (sin CH / sin AC)
or

ANGLE MADE BY THE VERTICAL 141
SQ = SS' × (R cos S'SQ / R) = (P × R sin ZS / R) × (R cos S'SQ / R)
¹= (P / R) √((R sin ZS)² - (R sin ZN')²)
²= (P / R²) × R sin N'S × R cos ZN', where N' is the nonagesimal.
Thus R cos S'SQ is seen to be
= (R sin N'S × R cos ZN') / (R sin ZS)
The Indian method is fully described by Bhāskara in his
'Golādhyāya. VIII, 12-25. The truth of the Indian rule for R cos
S'SQ is easily seen from the spherical triangle ΠZS, where Π is
the pole of the ecliptic.
Greek Method :
³Ptolemy takes SK and SL=90° each, along the vertical
circle ZSEK and the ecliptic N'SA. The great circle through K
and L cuts the horizon at R which is the pole of the vertical
circle. He takes SKL for the triangle and EAR for the trans-
versal, then
(sin SE / sin EK) × (sin KR / sin LR) × (sin LA / sin AS) = 1
or sin LR = (cos ZS × cos AS) / (sin ZS × sin AS)
or cos S'SQ = cot ZS × cot AS = tan SE × cot AS.
The Indian and the Greek rules are altogether different
both in form and method. There can, therefore, be no question
of any connection between them.
Problem VIII :—
To convert the Celestial Longitude of a Heavenly
Body into its Polar Longitude
If σ be the position of a (Fig.13), γK and σK are the celestial
longitude and the celestial latitude, respectively ; γM and σM
are the polar longitude and polar latitude ; γN and σV are the
right ascension and declination of the star.
Indian Method :
All Indian astronomers attempt at finding MK which, sub-

1. Āryabhaṭa, Gola, 34; Pañcasiddhāntikā, IX, 22 BrSpSi, XI, 23.
2. BrSpSi, V, 4-5; Sūryasiddhānta, V, 7-8 Bhāskara, Grahagaṇita, XII, 4.
3. Manitius, ibid, p. 119.