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

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.

142 GREEK AND INDIAN METHODS
tracted from, or added to, γK the celestial longitude, gives γM
the polar longitude.
According to Āryabhaṭa¹,
MK = (σK × R vers γK × R sin ω) / R² .
Brahmagupta² makes a dis-
tinct improvement on Āryabhaṭa
and gives his rule for finding the
projection MK on the celestial
equator.
If P be the celestial pole, PKH
the secondary to the equator,
Brahmagupta says that,
Fig- 13
NH = (σK × R sin (γK+90°) × R sin ω) / R
If from σ, σR is drawn perpendicular to PKH, it is evident
that,
R sin σ R = (R sin σ K × R sin σ KR) / R
According to Āryabhaṭa and Brahmagupta, as explained
before,
R sin σKR = (R sin (γK+90°) × R sin ω) / R
Hence Brahmagupta intends that,
NH = σR = (σK × R sin σKR) / R
which is rather a big assumption. He then directs the finding of
the part of the ecliptic of which σR or NH is the projection on
the equator thus approximately to MK.
Āryabhaṭa, Brahmagupta³ and the modern Sūryasiddhānta
take the declination σN = σK + KH where σK is small. They
do not consider the case where σK is large.
Bhāskara alone gives us fairly correct rules for this trans-
formation of co-ordinates.

1. Āryabhaṭa, Gola, 36.
2. BrSpSi X, 17.
3. BrSpSi, X, 15, Sūryasiddhānta, II, 58.

CELESTIAL AND POLAR LONGITUDE 143
In order to find σN, he would multiply
σK by (R cos σKP) / R ; according to him,
σN = (σK × R cos σKP) / R + KH¹
This is a decided improvement on Brahmagupta's corres-
ponding rule. The declination σN obtained would be very
nearly accurate.
Having obtained σN, Bhāskara² then directs the finding of
NH, thus,
NH = (σK × R sin σKP) / (R cos σN)
He then directs the finding of MK on the ecliptic of which
NH is the projection by means of the times of rising of the signs
of the zodiac on the equator.
Thus, the Indian methods show a beginning and develop-
ment only. The Greek method as given by Ptolemy is mathe-
matically accurate.
Greek Method³
To transform the celestial longitude and celestial latitude to
right ascension and declination.
Let the great circle ΠσK meet the equator at Δ. Ptolemy
would then from the given value of γK, find γΔ and ΔK by
using his tables for the rising of signs of the zodiac on the equa-
tor. He then takes ΠPσ for the triangle and γNΔ Q for the
transversal. The Menelaus' Equation, then, is
(sin ΠQ / sin QP) × (sin PN / sin Nσ) × (sin σΔ / sin ΔΠ) = 1²
Here ΠQ = 90° + ω, QP = 90°, PN = 90°; σΔ = σK + KΔ.
ΠΔ = 90° + KΔ, whence Nσ is obtained.
He next takes PNQ for the triangle and ΠσΔ for the trans-
versal,

144 GREEK AND INDIAN METHODS
It is almost needless to say that neither in the method nor
in the rules is there any agreement between the Indian and
Greek spherical astronomy in the solution of this problem.
Kaye’s view¹
As to Kaye, it appears that he has not been able to find a
method in the translation of the Sūyrasiddhānta by Burgess. The
figures of his paper referred to before do not show the “Akṣakṣet-
ras” even in their projections on the meridian place. He refers
to Braunmühl’s History of Trigonometry but does not appear to
have been able to follow him in his “Methode der indischen
Trigonometrie.” Kaye, however, is not slow in belittling Indian
trigonometry when he says :—The Indian astronomers employed
the sine function principally and the versed sine occasionally ;
they never employed the tangent function; and generally, but not
always, preferred to employ the sine of the complementary angle
rather than the cosine functions.’’
It is evident that Kaye
never understood the meaning
of the Indian functions of
‘sine’ and ‘cosine.’ These
functions are fully explained
by Bhāskara² when he
says :—
“Of that point the dis-
tance from the east-west line
is the sine and the distance of
the point from the north-south
line is the “cosine”.
[Fig. 14: Circle with points N, S, E, W; P₁, P₂, P₃, P₄; N₁, N₂, N₃, N₄; M₁, M₂, M₃, M₄, A]
Fig. 14
In (Fig. 14), of the arc AP₁, P₁M₁ is the “sine”
and P₁N₁ is the “cosine” of AP₂, P₂M₂ is the “sine” and P₂N₂
is the “cosine”; of AP₃, P₃M₃ is the “sine” and P₃N₃ is the
“cosine”; etc. It is evident that a better definition of these fun-
ctions was never given.
We have thus seen that some of the solutions of Āryabhaṭa

1. J. A. S. B., N. S., XV, p. 154.
2. तस्य बिन्दोः प्राच्य-परायाश्च यदन्तरं सादोर्ज्या ।
   बिन्दोर्याम्योत्तरायाश्च यदन्तरं सा कोटिज्या ॥
   —(Bhāskara, Grahagaṇita, commentary. II. 88-21)

KAYE'S VIEW 145
are imperfect, of Brahmagupta the solutions are more accurate,
while those of Bhāskara are generally mathematically correct.
The date of the scientific ancient Indian Astronomy is indeed
499 A. D., while that of the Syntaxis is about 150 A. D. It is by
these shortcomings and differences in the methods, new ideas (e.g.,
the idea of the differential calculus)* and the like, that we
can safely say that Indian Astronomy in its scientific form,
although of a later date than the "Syntaxis" of Ptolemy, is origi-
nal and not borrowed from foreign source. There is evidence
that some crude form of Greek astronomy was transmitted to
India and went by the name of the "Romaka" or the "Pauliśa"
Siddhānta, prior to the time of Āryabhaṭa but our great Indian
astronomers, Āryabhaṭa with his pupils, Varāha-mihira and
Brahmagupta, had to construct a new science altogether.
(This Chapter is almost a reproduction of the paper by
P. C. Sengupta, as acknowledged earlier).
—: o :—
Reference
P.C. Sengupta : The Khaṇḍakhādyaka, 1934

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