पञ्चसिद्धान्तिका (आचार्य वराहमिहिर - सूर्य, रोमक, पौलिश, वासिष्ठ एवं पितामह सिद्धान्त)
Panchasiddhantika of Acharya Varahamihira with Commentary
आचार्य वराहमिहिर द्वारा
पृष्ठ 47, कुल 419 में से
संदर्भ में पढ़ेंI.21 I. INTRODUCTION OF THE WORK 21 Lord of the Horā 20. Take the remainder set apart in verses 17-18. Divide out by 7 and the remainder is the Lord of the Day, counting from the Sun. Take this remainder, multiply by 3, add 1, and add also the number of horās (i.e. the hours) counted from the beginning of the day, (i.e. the previous sunset) inclusive of the horā in which the taken moment falls. Multiply by 5 and divide out by 7. The remainder, counted from the Sun, gives the Lord of the Horā. If the Lord of the day is dth from the Sun and the time taken falls in the hth horā, then the number for the Lord of the Hora is (3d + 1 + h) × 5. It should be noted here that the horā, h, is counted from sunset, because the time of Epoch is sun- set and the day is said to commence there. The derivation of the two rules: The rule for the Lord of the Day is obvious for the order of the Lords, Sun Moon, Bhauma, etc. is meant to be the order of the Lords of the weekdays, Sunday, Monday, etc. The rule for the Lord of the horā is derived thus: From the Śāstra we learn that the Lord of the horā beginning at sunrise is the same as the Lord of that day. The Lord of the horā begin- ning Sunday, i.e. of the horā just after sunset of Saturday, (i.e. Mandavāra), is Budha, since the Lord of the horā after sunrise on Mandavāra is Manda and the successive Lords of the horās are the fifth after each, i.e. the sixth counting from each. (vide the next verse, 21). Budha is the 4th in order. After this if (n − 1) horās are gone, the Lord of the nth horā is given by (n − 1)5 + 4. Let us find the Lord of the horā for the h-th horā of the d-th day. This is {(d − 1)24 + h}th horā. Therefore the Lord of the horā is, substituting this for n in the above formula, {(d − 1)24 + h − 1} 5 + 4 = (24d + h − 25) 5 + 4 = (21d + 3d + h + 1 − 26) 5 + 4 = (3d + 1 + h)5 + 4 + 5 × 21d − 5 × 26 = (3d + 1 + h) 5 + 105d − 126 = (3d + 1 + h)5 + 15d × 7 − 18 × 7. As no change in the Lord happens by adding or deleting multiples of 7, this reduces to (3d + 1 + h)5, which is the rule given. (Here too the deri- vation of M.M. Sudh. is wrong. Let the readers examine his commentary.) The acceptance of the expression vyeka in place of the ms. reading 'dhyeka both by TS and NP has rendered their trans- lations incorrect. Example 7. (a) Who is the Lord of the Day, for the day given in Ex. 5? (b) On the same day, who is the Lord of the Hora, fifth after sunrise? (a) The remainder set apart according to verses 17-18 is 666. Dividing out by 7, the remainder left is 1, i.e. the Lord of the Day is the Sun. (b) In the example, d = 1, h = 5 + 12 = 17 (because h is counted from the beginning of the day, i.e. the previous sunset). Substituting, (1 × 3 + 1 + 17)5 = 105. Casting out 7, the remainder is 0 or 7 and the 7th from the Sun, Manda is the Lord of the horā. वर्षाधिपश्चतुर्थो मासाधिपतिस्ततो योऽन्यः । होराधिपश्च षष्ठो निरन्तरं दिवसनाथश्च ॥ २१ ॥ 20. Quoted by Utpala on BS 2, p.34. c. C.D. U. पञ्चम; 20a. B1.2. सप्तोद्धृते B1. सप्तहृते; B2. सप्तहृते; C.D. U. सप्तहतो b. A1.2. B1.2. ०ध्येकशहोरादिः; C.D. U. त्रिगुणो d. A1.2. विज्ञेया; B1. विज्ञेय व्येको युतश्च होराभिः A1.2. कालहोरेशाः; B1. कायहोरेशः; B2. कायहोरेशः