पञ्चसिद्धान्तिका (आचार्य वराहमिहिर - सूर्य, रोमक, पौलिश, वासिष्ठ एवं पितामह सिद्धान्त)

IX.20 | IX. SAURA-SIDDHĀNTA — SOLAR ECLIPSE | 219
मध्यज्याकृतिविश्लेषितां पृथक् स्थाप्य मूलमेकस्याः ।
सवितुर्दृक्शेपाख्यं संस्मृत्यर्थं पृथक् स्थाप्यम् ॥ २० ॥
Dṛkṣepa of the Sun
19. Find the sine of the longitude of the Orient Ecliptic Point (o.e.p.) at new
moon, multiply by the sine of maximum declination (of the Sun, 48′ 48″) and
divide by the sine of the colatitude. (This is sine amplitude of o.e.p., called
Udayajyā.) Multiply this by the sine of the zenith distance (z.d.) of m.e.p.
already found, and divide by 120′. Square the result, and subtract from the
square of the sine z.d. of the m.e.p.
20. Set the remainder in two places. In one place, find its square root. This
is the sine of the zenith distance of the nonagesimal (z.d. of n.) called the Sun's
dṛk-kṣepa. Keep this safe aside for future work.
The following is to be done:
(i) Using the vināḍīs of ascensional differences of the place, the o.e.p. at new moon is to be
found.
(ii) Sine amplitude of o.e.p. = sine z.d. of m.e.p. × 48′ 48″ ÷sine colatitude.
(iii) Sine (m.e.p. ~ nonagesimal) = sin z.d. of m.e.p. (ii) ÷ 120.
(iv) Square of sine (z.d. of n) = (sine z.d. of m.e.p.)² – (iii)².
(v) Sine z.d. of n = √(iv).
Example 13. Continue example 10, given already lat = 10° 24′ and new moon is at nā. 20-40.
(i) Let us take it that using the ascensional differences of the place (given by chap. IV), the o.e.p.
found is rā. 5-27-56.
(ii) Sine amplitude of o.e.p. = sine rā. 5-27-56 × 48′ 48″ ÷ sin (90° – 10° 24′)
= sin 2° 4″ × 48′ 48″ ÷ sin 79° 36′
= 4′ 20″ × 48′ 48″ ÷ 118′,0 = 1′ 48″.
(iii) Sine (m.e.p. ~ ṇ) = 28′ 9″ × 1′ 48″ ÷ 120 = 25″.
(iv) Sin² (z.d. of n) = (28′ 9″)² – (25″)² = 792′ .25.
(v) Sin (z.d. of n) = √792.25 = 28′ 9″.
19a. A.B. विलग्ना ज्या | b. B. ०षितां; C.D. ०षिता. A.B.D. स्थाप्या; C. स्थाऽपि.
b. A. काष्टांत; B3. कापांत० | A.B. ०मेकस्या
c. B. मध्यमज्याघ्नी | c. A. ०दृक्षेपाख्यं; B. ०दृक्क्षेपाख्यं
20a. B1. ०तति०; B2.3. ०तति०. D. ०विशे० | d. B2. पृथक्थो य ||