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

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

177
In the beginning of the Triprasnādhyāya, Bhāskara says
“साक्षे देशे खगोलवलयानां, तिरश्चीनभगोलवलयानां च संपातात्त्र्यस्राणि
क्षेत्राण्युत्पद्यन्ते, तान्यक्षक्षेत्रसंज्ञानि” ie. In a place having a
latitude the diurnal paths of stars and planets will be
inclined to the fundamental circles of the celestial sphere
namely horizon, meridian and prime vertical and so their
intersection gives rise to what are called latitudinal tri-
angles, in which the angles would be ϕ, 90-ϕ and 90
where ϕ is the latitude. Thus in fig. 21 the projections
of the triangles ESB, EDB, DSB, EDF, EBF, ESF and
EQG on the meridian plane will be all triangles in which
the angles will be ϕ, 90-ϕ and 90° so that they are all
latitudinal triangles. These are all similar to the funda-
mental gnomonic triangle gmn of fig. 18 wherein also the
angles are ϕ, 90-ϕ and 90°. Here, there is one important
point to be observed. We have said “their projections
are all similar”. In fact the corresponding spherical
triangles enumerated above are all apparently similar,
though they are not viewed in modern astronomy as
regular spherical triangles, because all the three sides of
the triangles are not arcs of great circles. It is not possi-
ble to apply Napier’s rules to these triangles for the reason
mentioned above. None the less, the property of simil-
arity of the projected right angled triangles is made use
of in Hindu Astronomy to obtain the magnitudes of the
sides of the projected triangles.
EW is called the prāk-pratīchī sūtra, SS′ the Uda-
yāsta sūtra; similarly if lines through F, B, D, A parallel
to EW be drawn, these sūtras will intersect the meridian
planes in points which constitute the projected triangles
mentioned above. Let us study these triangles which we
connote by the same letters with lowered indices. Thus
for example E₁ S₁ B₁ is the projected triangle of ESB
where of course E₁ will be the centre of the celestial
sphere. The lines drawn parallel to SS′ or EW through
B, D etc. will be denoted as BB′, DD′, FF′ etc.
23

178
In the triangle E₁ S₁ B₁, E₁ S₁ is called Agrajyā, S₁ B₁
Kujyā, and E₁ B₁ Krāntijyā. Of the sides of ESB. ES
and EB are arcs of great circles whereas SB is the arc of
a small circle. In the projected triangle E₁ S₁ B₁, E₁ S₁
will be equal to the perpendicular from S on Eω, which
will be H sin (ES) and is called Agrajyā ; E₁ B₁ will be
equal to the perpendicular from B on Eω which will be
H sin BE and as such called Krāntijyā. But S₁ B₁ which
is equal to the perpendicular from S₁ on BB' the diameter
of the diurnal circle is the H sine of SB in the diurnal
circle and is called Kujyā. It will be noted that the per-
pendiculars from S on Eω, and BB' and the perpendicular
from B on Eω do not form a triangle by themselves but
by the theorem of three perpendiculars, if SL be the
perpendicular from S on the plane of the unmandala ie.
the great circle EBP, and if LM be perpendicular from
L on Eω, SM will be perpendicular on Eω. Here the
perpendicular SL will be the Kujyā, and SM the Agrajyā,
whereas LM is not actually the H sine of BE but is equal
and parallel to it.
Similarly take the spherical triangle SAE. This is a
regular spherical triangle because the three sides are arcs
of great circles. Napier's rules can be applied to this
triangle and we have the formula sin SA = sin SE sin
SÊA or in Hindu form H sin SA = (H sin SE × H sin Ê) / R
or RH sin δ = Agrajya × H cos ϕ ie. Agrajyā = (RH sin δ) / (H cos ϕ) II
Again sin EA = tan SA × tan Ê where H sin EA is
called Charajyā and tan E = cot ϕ so that Charajyā =
tan δ tan ϕ in modern form, and the Hindu form is
<u>R tan δ tan ϕ</u>

179
From I H cos δ / R = Kujyā / Charajyā so that
Kujyā = H cos δ / R × R tanδ tanϕ = H sinδ tanϕ III
In Triprasnādhyāya, the elements of all the eight
latitudinal triangles are found by using their similarity
with the fundamental gnomonic triangle. The important
elements that will enter into computation are (a) Charajyā
(b) Kujyā (c) Agrajyā (d) Taddhriti ie. S₁ F₁ the proje-
ction of SF on the plane of the meridian (e) Sama-Sanku
= E₁ F₁ = H sin EF (f) Krāntijyā = E₁ B₁ = H sin EB
= H sin δ (g) Lambajyā = H sin QS = H cos ZQ =
H cos ϕ (h) Akshajyā = H sin ZQ = H sin ϕ (i) B₁ D,
= Ud-Vritha-Sanku = H cos ZB (j) Dinārdha-Sanku =
H cos Zq. The following points will be noted.
(i) In the triangle E₁ Q G₁, the projected triangle
of EQG on the meridian plane, noting that E₁
is the centre of the celestial sphere E₁ Q = R,
QG₁ = H cos ϕ and E₁ G₁ = H sin ZQ =
H sinϕ.
(ii) S₁f₁ = H sin SB + H sin fB (both the H sines
pertaining to the diurnal circle.
(iii) Sama Sanku is the H cosine of the Zenith-
distance when the Sun or celestial body is on
the prime-vertical.
(iv) BD is an arc of the great circle ZB, so that the
Unmandala Sanku is the H cosine of ZB.
(v) Dinārdha — Sanku = H cosine ZQ.
(vi) These Sankus are the H sines of altitudes or
H cosines of Zenith-distances and they are in
the planes of the respective great circles.

180
(vii) Lambāṁsa-chāpa is the arc of the colatitude
QS where S is the South point so that Lam-
bajyā is the H sine of QS or H cosine of ZQ.
This Lambajyā will be seen to be the diameter
of a small circle parallel to the Equator and
passing through Z.
Verses 49-51. A different method of obtaining
chara.
This chara can be had by the so-called chara-segments
of the locality using a process similar to that of finding
the H sines of the smaller table of nine H sines using
λ/3 where λ is the Sāyana longitude of the Sun (i.e. The
Hindu longitude plus the arc of ayanāṁśas is called the
Sāyana longitude or the modern longitude measured from
v along the ecliptic to the Sun). Find the charas of 30°,
60° and 90° of the ecliptic measured from v. Subtract the
first from the second, the second from the third. Thus we
have C30°, C60°-C30°, C90°-C60°, (where Cθ° signifies the
chara of θ°) which are called chara-Khandas or chara-seg-
ments. The equinoctial shadow multiplied by 10, 8, 3⅓
gives the approximate values of the chara-segments in
Vinādīs (a sidereal day is divided into 60 nādīs and each
nādī consists of 60 Vinādīs). The chara-segments thus
measured in Vinādīs are rather approximate. If further
exactitude is required, better take the arc in units each
of which rises in ⅙th of a Vinādī (This ⅙th part of a
Vinādī is known as a prāṇa i.e. the duration of the
interval between two inhales of a healthy person reckoned
as 4″ of time).
Comm. The charas of 30°, 60° and 90° of modern
longitude are the values of the arc EA, (Ref. Fig. 21)
when S has longitudes 30°, 60° and 90°. We have the
folmula
Charajyā = R tan δ tan ϕ

181
Taking the equinoctial shadow equal to 1 Angula
means tan ϕ = 1/12. As charajyā is proportional to tan ϕ,
for any equinoctial shadow of s angulas, tan ϕ being
equal to s/12 the charajya got above is to be multiplied by
S only to obtain the charajyā in any place where the equi-
noctial shadow is s angulas. Putting the modern longitu-
des equal to 30° & 60°, if the corresponding declinations
be δ₁, δ₂ sin δ₁ = sin 30 sin ω, sin δ₂ = sin 60 sin ω. Tak-
ing ω = 24° and applying logarithmic tables
log sin δ₁ = 9.6990 + 9.6093 = 9.3083 so that δ₁=11°–44′
log sin δ₂ = 9.9375 + 9.6093 = 9.5468 so that δ₂=20°–38′
Now from the formula for charajya cited above viz.
H sine (chara) = R tan δ tan ϕ or sine (chara) =
tan ϕ tan δ, putting tan ϕ = 1/12 and applying tables,
using the values of δ got above,
charajya for 30° = (tan 11°–44′) / 12
and charajyā for 60° = (tan 20°–38′) / 12 so that
log (sine chara) = 9.3175 – 1.0792 for 30° and for 60°
log (sine chara) = 9.5758 – 1.0792
∴ Chara for 30° or C (30)° = 59′ and C (60°) = 1° – 48′
Converting these arcs into their rising times at the rate of
6′ per Vinadi, we have C (30°) = 10, and C (60°) = 18
Noting δ₃ = ω, sin (chara) for 90° = (tan 24°) / 12 so that
log sin (C 90°) = 9.6486 – 1.0792 = 8.5694 so that
C (90°) = 2°–8′ = 21⅓ Vinādis.
Thus (C 30̄) = 10, C (60) – C (30) = 18 – 10 = 8
C (90°) – C (60°) = 21⅓ – 18 = 3⅓ so that the chara
Segments are respectively 10, 8, 3⅓ as given by Bhāskara.

182
For a given place of equinoctial shadow equal to s″, we
have to multiply 10, 8, 3⅓, by s, which will be the chara-
segments for the place.
The meaning of the first half of the verse 49 is as
follows.—Let the equinoctial shadow for a place be 3
Angulas. Then the chara-segments for that place are 30,
24, 10. These are three in number for a longitude λ of
90°. If the longitude be 44° (say) then proceed as we have
done to find sin 24°, using the method of Bhogya Khanda
Sphutīkaraṇa with respect to the table of H sines namely 21,
20, 19, 17, 15, 12, 9, 5, 2. Proceeding as directed the Sphuta
Bhogya Khanda for 14° is ( (3 × 14) / 30 = 7/5 ; 30 - 7/5 = 28⅗ ;
(14 × 28⅗) / 30 = 2002 / 150 = 13⅓ ; 30 + 13⅓ = 43⅓ ).
Proceeding according to the modern formula we have
Charajyā = tan δ tan ϕ where ϕ = 3/12, and sin δ =
sin 44° sin 24°
log sin δ = 9.8418 + 9.6093 = 9.4511; δ = 16°-25
log tan δ = 9.4693 ∴ log sin (chara) = 9.4693 + log tan ϕ
= 9.4693 + log 3/12 = 9.4693 - .6021 = 8.8672
∴ Chara = 4°-14' = 254 / 6 = 42⅓ Vinadis whereas we have
got by the Hindu method 43⅓ which is near the truth.
Verse 52. To find the durations of day and night.
Fifteen ghatis increased or decreased by the Chara-
nadis, according as the Sun is in the northern hemi-
sphere or southern, gives half the day of the locality and
the difference of 30 nādis and the above half-day gives
half the duration of night.
Comm. Ref. fig. 21. Let S be the Sun rising in the
northern hemisphere when his declination is north. Then

183

from the figure ÂPQ is the rising hour-angle = ÂPE +
ÊPQ = ÂPE + 90° where 90° correspond to 15 nādīs. So
we have to add ÂPE expressed in nādīs equal to the rising
time of AE. Let us find the duration of the day for the
place where s=3″ and when λ of s=44° ; the latitude of
the place will be 14° (from tables) when λ 44°, we have
found above that 42-20 Vinadis is the chara expressed in
time. Hence half the day = 15-42-20 or duration of day
31-25 ; duration of night = 28-35.
Note (1) In modern astronomy we have the formula
cos h = — tan ϕ tan δ where h is the rising hour angle.
Putting h = H + 90° where H stands for the arc EA of
fig. 21 cos (90+H) = — sin h = — tan ϕ tan δ so that
sin H = tan ϕ tan δ ie. R sin (Chara) = R tan ϕ tan δ
which accords with the formula found before. The word
chara used for EA means etymologically रविसञ्चारवशेन दिन-
प्रमाणे विकारः i e. the variation in 15 ghatis of the eqiunoctial
half-day on account of the Sun's variation in declination,
(2) If ϕ = 0, Chara = 0 so that duration of half-
day is 15 ghatis ie. on the terrestrial equator, whatever
be the Sun's declination, the length of the day will be
always 12 hours.
(3) Let δ = 0 so that chara = 0 ie. whatever be
the latitude (provided ϕ ≯ 90-δ as we shall see shortly).
the day and night will be each of 12 hrs.
(4) Let δ be negative, so that the arc connoting
chara ie. EA will be above the horizon, and consequently
the duration of half-day will be less than 6 hrs. by the
time that is taken for EA to rise. This can be seen other-
wise also as cos h = — tan ϕ tan δ = + ve so that h < 90°
which means half-day is less than 6 hrs.
(5) We could also treat the case when ϕ is negative,
but as this case is not in the purview of Hindu Astro-
nomers who had only India in their mind and as such
were concerned primarily with positive latitudes. If, how-
ever, we consider a negative latitude, when δ is + ve, cos h
will be positive and if δ is — ve, cos b will be negative.
This means that when the Sun is in northern latitudes, the
southern latitudes will have their day less than 12 hours
and when the Sun is in southern latitudes, their day will
be greater than 12 hrs.

184
Fig. 22
(6) Let ϕ + δ = 20°. (Ref. fig. 22) ie. imagine
the Sun to rise at N, the north point so that RN + NP =
δ + ϕ = 90°. Then the Sun's diurnal path will be entirely
above the horizon, which means that what is called
'perpetual day' begins for that place on that day and lasts
as long as δ ⩾ 90 - ϕ. For the same place, let δ =

• (90 - ϕ) = QS. In this case the Sun sets at S, and
  what is called perpetual night begins and lasts till the
  southern declination of the Sun is greater than 90 - ϕ.
  The duration of perpetual day can be found as follows
  which applies to the perpetual night as well.
  Fig. 23
  (Ref. fig. 23). Let A be the point at which the
  declination is 90 - ϕ; let S be the summer solstice and let
  B be the point where again the declination is equal to
  90 - ϕ. So long as the Sun traces the arc AB = 2 AS =

185
(90-rA), there will be perpetual day. But sin δ = sin λ
sin ω so that sin λ
= sin δ / sin ω . Putting δ = 90 - ϕ, sin λ = cos ϕ / sin ω ;
Sin λ = sin γA = cos AS ∴ cos AS = cos ϕ / sin ω
∴ 2 AS = 2 cos⁻¹(cos ϕ / sin ω). Supposing AS expressed
in degrees and assuming the Sun goes along the ecliptic
with uniform motion, since he takes 365¼ days to trace
360°, to trace 2 AS, he takes (2 AS / 360) × 365¼ days = 365¼ / 180
× cos⁻¹(cos ϕ / sin ω) which is the length of the perpetual day.
Verse 52. The correction known as Chara.
The daily motion of the planet being multiplied by
the chara expressed in asus and divided by the asus in a
day viz. 21659 and the result being subtracted from or
added to the planetary position at Sunrise according as
the Sun is in the northern or southern hemisphere. The
result is to be added to or subtracted from the planetary
position at Sunset.
Comm. The mean planets computed hold good at the
Sun-rise at Lanka i.e. at zero latitude; they have to be
converted to hold good at the local Sun-rise. In other
words in fig. 21, the mean planet computed is B which is
on the Lanka horizon, whereas we have to get S, the same
planet on the local horizon. The position of S is earlier
than that of B by the time interval indicated by the arc
SB which is measured by EA the chara because the
position B on the Lanka horizon is later than the position
S on the local horizon. If the mean planet moves in a
day of 21659 asus by the arc denoting its daily mean
motion, by how much does it move in the time of chara
expressed in asus? The result is
(Chara in Asus × mean daily motion) / 21659 .
24

186
This result is to be subtracted from the position of B
to get the position of S. If the position B′ which is the
setting position at Lanka, and if we have to get S′ the
local setting position, in as much S′ is later than B′ by
the same arc, we have to add the above result to the posi-
tion B′ to get S′. Here the asus in a day is given to be
21659 and not 60 × 60 × 6=21600 because the mean daily
motion is during the course of the Sāvana day i.e. the true
solar day and not during the course of a sidereal day of
21600 asus. The Sāvana day exceeds the sidereal by the
time taken by the arc moved by the Sun during that day
which is very approximately 59′. Since a minute of arc
of the Equator rises in an asu i.e. in 4″, so 59′ of the Sun's
motion is covered in 59 asus which is to be added to 21600
asus of the sidereal day to get the Sāvana day approxi-
mately. It will be noted that this correction of chara in
the planetary positions is due to latitude of the place and
if the latitude is zero, it need not be done. Also if δ = 0,
it need not be done, for, then, the Sun or the celestial
body whose δ = 0 will be rising at E itself in which case
the chara EA will be zero.
Verses 54, 55. The H sines of 30°, 60° and 90°, being
squared and decreased by the squares of their respective
declinations, the square-root of the differences being taken,
and the result being multiplied by the radius, is to be
divided by the respective H cosines of the declination.
The first of the three results, the difference of the second
and the first and the difference of the third and the second
will give us the rising times of what are called the Sāyana
rasis of Mesha, Vrishabha and Mithuna; their reverses
will then give the rising times of the next three; then
the original ones those of the next three and again then
reverses those of the last three.
Comm. There are twelve Rasis in the Zodiac of equal
interval. Measuring from r, and taking arcs of the

187
ecliptic successively each of 30°, we have what are called
Sāyana Rasis or Rasis taking the Ayanāṁsa into consi-
deration or in other words measuring from r. On the
other hand successive arcs each of 30° measured from the
zero point of the Hindu Zodiac, constitute what are called
the Nirayana Rāśis. The words Meṣa, Vṛṣabha etc.
signifying the shapes of the constellations apply strictly
to the Nirayana Rāśis. But nonetheless, by the conven-
tion what is called Upachāra, we name the Sāyana Rāśis
also by the same names. In as much as the zero-points
of the Hindu Zodiacs is ahead of the modern zero-point by
an arc which is the arc of Ayanamsa or accumulated pre-
cession, the words Sāyana and Nirayana came into vogue.
Once upon a time approximately in Varāha’s time or
rather 499 A. D, the two zero-points coincided and then
the Sāyana and Nirayana Rāśis were the same. Gradually
on account of the phenomenon of precession r preceded,
and today the distance between r and the Hindu zero-
point is about 20°–30′. Of late, there has been a big
controversy as to what exactly the arc is and the Calendar
reform committee has adopted a value far more than what
could be justified. The present author opines, that we
have no right to set aside a statement of no less an astro-
nomer than Varāhamihira who stated explicity ‘आश्लेषार्धात्
दक्षिणमुत्तरमयनं रखेर्धनिष्ठाद्यम् , नूनं कदाचिदासीत् येनोक्तंपूर्वशास्त्रेषु,
साम्प्रतमयनं सवितुः कर्कटकाद्यं मृगादितश्चाऽन्यत् उक्ताभावो विकृतिः
प्रत्यक्षपरीक्षणव्यक्तिः” i.e. True it is that once upon a time,
the Sun began his southern journey when he was mid-way
the constellation of Asleṣa and his northern when he was
at the beginning of Dhaniṣtha, because it was stated in
ancient texts. (The allusion is to Vedāṅga Jyotiṣa
where it was stated as such); but now the southern journey
of the Sun begins from the point marking ¾ of the con-
stellation of punarvasu which is the beginning point of
the Karkaṭa Rāśi and his norther from the beginning of
Makara i.e. from the point marking ¼th of the constellation
of Uttarāṣādha; there has been a change from what was

188
stated in the ancient texts ; let people verify this by actual
observation". Accepting an Ayanāṁsa which goes
against the statement of this great astronomer, who said
that he observed and called upon others to observe, is
really unwarranted, especially when the adopted Ayanāṁsa
of the Calendar reform committee goes on the basis of
surmises and consensus. We shall deal with this topic
in further detail in an appendix to this work, because it
is a really important issue and has been wrongly solved.
The importance of knowing the exact value of the arc is
clear when we observe that from the correctly computed
planetary positions of modern astronomy this ayanāṁsa
is being subtracted to give the positions measured from
the Zero-point of the Hindu Zodiac. An error in the
Ayanāṁsa therefore vitiates the positions obtained by the
above method.
Coming to the point, the problem on hand is to
obtain the rising times of the Sāyana Rāśis at a place of
zero latitude i.e. what are called Laṅkōdaya times of
Sāyana Rāśis (Ref. fig. 21). Let rS = 30° so that rA the
equatorial arc gives the rising time of rS. We could have
the magnitude of this arc by the formula derived from
Napier's rules namely cos ω = tan α tan λ I.
But as in Hindu trigonometry the tangent functions of
angles are not used, Bhāskara gave the following formula
Sin α = √(sin² λ − sin² δ) / cos δ II. This formula could be
derived easily from formula I and the formulae cos λ =
cos α cos δ III and sin δ = sin λ sin ω IV ; for multi-
plying the right-hand sides of I and III we have cos ω
cos λ = (sin α cos δ) / tan λ i.e. sin λ cos ω = cos δ sin α
∴ sin α = (sin λ cos ω) / cos δ V. But from IV sin ω =

--- ------------------ = ------------------- sin λ cos δ cos δAnd then:H·cos δ as stated.? Wait, where does H·cos δ as stated.come from? Wait! Look at line 5:H sin α = R √(H sin² λ - H sin² δ) / (H cos δ) ...Wait, is line 5:H sin α = R √(H sin² λ - H sin² δ) / (H cos δ)Wait, why is thereH·cos δ as stated.? Could it be: ... / H·cos δ as stated.? Wait! Look at the vertical alignment: Line 5: H sin α = R[fraction numerator:√(H sin² λ - H sin² δ)] Wait, what is the denominator of line 5? H cos δ! Then on line 6: sin λ × 1/sin λ ...Wait, why isH·cos δ as stated.BELOW line 6? Wait, look at the line aboveH·cos δ as stated.: There is a horizontal line above H·cos δ: H̄·cos δ as stated.? Wait! Is that H·cos δ as stated.or is that a fraction line? Wait, look at the letter beforecos δ`:
It has a bar above it:

190
From M drop a perpendicular ML on r ♎ so that SL will
be perpendicular from S on r ♎ by the theorem of three
perpendiculars. SL = H sin λ. Also ML will be equal
to the perpendicular from S on the diameter of the diurnal
circle of S parallel to r ♎ so that ML = SN = H sine of
the arc in the diurnal circle corresponding to Kr
∴ SW² = SL² - LN² = H sin² λ - H sin² δ
∴ SN = √(H sin² λ - H sin² δ)
∴ The length of the perpendicular from K on r ♎ =
R × √(H sin² λ - H sin² δ) / (H cos δ) since corresponding lines of
the diurnal circle and the equator stand in the ratio of
H cos δ : R (Vide fig. 20). But this ⊥ᵃʳ is H sin α =
R √(H sin² γ - H sin² δ) / (H cos δ) .
If α₁, α₂, α₃ be the Right ascensions of the points on
the ecliptic whose modern longitudes are 30°, 60° and 90°,
expressed in asus, then α₁, α₂ - α₁, α₃ - α₂ will give the
rising times of the arcs of the ecliptic which stand for
Sāyana Meṣa, Sāyana Vriṣabha and Sāyana Mithuna.
The rising times of the next three Rasis will be the same
in reverse order since Karkata is symmetric with Mithuna
with respect to the Equator and similarly Simha and
Kanya symmetric with Vriṣabha and Meṣa. The next
three are again symmetric with Meṣa, Vriṣabha and
Mithuna and the last three with Mithuna, Vriṣabha and
Meṣa.
Verse 56. The H cosines of the ends of the Rasis
Karkata etc., being multiplied by the radius, and divided
by the H cosines of their respective declinations, and the
arcs of those H cosines being taken, subtract as before the
preceding from the succeeding. Then we have the rising
times of the Rasis beginning with Karkata.

191
Comm. This is clear from fig. 24. If S be the end of
Vṛiṣabha, S' in the figure denotes the end of Karkaṭa
them from the right-angled triangle SAP,
^
sin S'A = sin SA = sin APS × sin PS = sin A'K × cos SK
sin S'A
∴ sin A'K = ————— . But sin S'A = cos S'♈
cos SK
and cos SK = cos δ where δ is the declination at S.
A'K converted into time gives the rising time of AS'
ie. Karkaṭa.
cos S'
∴ the rising time of Karkaṭa = ————
cos δ
In tbe Hindu form, it will be
Karkaṭa-anta-Koṭijyā
Karkaṭa-Rāsi-Udayakāla = ————————————— × R
Karkaṭa-anta-Dyujyā
as stated.
cos ♈S
In modern terms sin A'K = cos ♈K = ————
cos δ
from the formula cos λ = cos α cos δ. Thus, virtually
the formula is a statement of the formula cos λ = cos α
^
cos δ. Aslo the formula sin AS' = sin P sin PS is
parallel to the formula sin δ = sin λ sin ω which we
proved already from Hindu methods.
Verse 57. Still an alternative method.
The H sines of Meṣa etc. being multiplied by
H cos ω divided by their respective H cos δ's and the
arcs thereof being subtracted as before the preceding from
the succeeding we have the rising times of Meṣa etc.
Comm. From figure 24,
^ SN
SN = SL cos LSN and ————— = perpendicular from
cos SK
K on ♈

192
∴ (SL cos ω) / (cos δ) = H sin rK. SL = H sin rS
∴ H sin rK = (H sin rS × H cos ω) / (H cos δ)
We proved the above in a modern way. The Hindu
concept is derived from the similarity of SML and ACM′
where C is the centre of the sphere and M′ is the foot of
the perpendicular from A on the plane of the equator
∴ LM / CM′ = SL / CA ∴ LM = (H cos ω × H sin rS) / R
Since LM = SN.
LM divided by H cos δ and multiplied by R gives H sin rK
∴ H sin rK = [(H cos ω × H sin rS) / R] × [R / (H cos δ)] =
= (H cos rS × H cos ω) / (H cos δ)
i.e. H sin α = (H sin λ × H cos ω) / (H cos δ)
Here H cos ω is called Trigṛha-dyu-maurvī because
it is the H cosine of the declination of λ when λ = 90°.
Verses 58, 59. The magnitudes of the rising times.
Those rising times are 1670, 1793, 1937 ; these in the
same and reverse orders diminished or increased by their
respective Chara segments which are also in the same and
reverse orders give the rising times of the Sāyana Rasis
beginning from Meṣa for the locality. The Rasis from
Tulā are in a reverse direction i.e. as the Meṣa is proje-
cting upwards above the horizon, Tulā will be projecting
below the horizon so that, the time taken by Meṣa to
rise is exactly the time taken by Tulā to set.
Comm. We shall compute the rising times of
Sāyana Rasis for Lanka first i.e. for zero latitude using
modern methods from the formula tan α = cos ω tan λ

198
log tan α = log cos ω + log tan λ; Put λ₁ = 30, and λ₂ = 60
and take ω = 24°; Let the corresponding α's be α₁, α₂
log tan α₁ = log cos 24 + log tan 30° (1)
log tan α₂ = log cos 24 + log tan 60° (2)
log tan α₁ = 9.9607 + 9.7614 = 9.7221
∴ α₁ = 27° - 48'
log tan α₂ = 9.9607 + 10.2386 = 10.1993
∴ α₂ = 57° - 42'
At the rate of 1 asu for 1', α₁ = 1668 asus α₂ = 3462;
α₂ - α₁ = 1794 and since α₃ = the right ascension of 90°
Longitude = 90°, α₃ = 5400 so that α₃ - α₂ = 1938.
These are given by Bhāskara as 1670, 1793, 1937, the first
exceeding by 2 asus, the second and third each less by
one asu and the total according with the total. The rising
times of Karkaṭa etc. will be 1937, 1793, 1670, 1670, 1793,
1937, 1937, 1793, 1670 respectively.
Let us then find the rising times of these Sāyana
Rasis at a locality say of latitude 13°. Refer to fig. 21.
Let rS represent Meṣa so that the rising times of rE is
equal to that of rS. But rE = rA - AE. We have seen
rA = 1670 using Bhāskara's value. Sin EA = tan 13°
tan δ₁ where δ₁ is the declination of S where rS = 30°.
Sin δ₁ = sin 30° sin 24°.
log sin δ₁ = 9.6990 + 9.6093 = 9.3083
∴ δ₁ = 11° - 44'
∴ log sin EA = log tan 13° + log tan 11° - 44 =
9·3634 + 9.3175 = 8.6809
∴ EA = 2° - 45' ∴ rE = 1670 - 165 = 1505 asus.
Similarly for λ = 60°, putting α₂, δ₂ in the place of
α₁, δ₁ and proceeding as before rE = rA₁ - A₁E; rA₁ =
3463; sin EA₁ = tan 13° tan δ₂ sin δ₂ = sin 60° sin 24°
25