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

XVIII.79 XVIII. STAR-PLANETS 363 Moreover, their filling the lacuna by seṣuḥ, meaning 5° is quite inadequate to make up zero. I have emended saṣṭāṣṭakena into pañcāṣṭakena, meaning 5 × 8 = 40, which will fit the number of days. Also, 10° synodic motion there requires 40 days and it is also the period from setting to going into superior conjunction. ṣaṣṭa is patently wrong spelling, and ṣaṣṭāṣṭakena is meaningless. But TS and NP keep it, which is wrong. That this is the segment of heliacal setting to conj. can be inferred from udayati given for the next segment, and 10° for heliacal setting and rising at superior conj. is given by many siddhāntas. The other minor emendations are TS's. Note 2. 1,18,122 seems to be a very large subtractive constant, equal to more than 300 years, while all others are very near VM's time. But I cannot think of any other number to fit, Note 3. The maximum elongation is seen to be 45°, correctly. (Cf. Table).

	1	2	3	4	5	6	7	8	9	10	Total
Degrees passed	– 9°<br>vakrāsta	– 21°<br>vakra	– 15°	+ 35°	+ 10°<br>asta	+ 10°<br>asta	+ 35°	– 15°	– 21°<br>asta	– 9°<br>vakrāsta	0
Given days	5	15	64	164	40	40	164	64	15	5	576
Correct days	5	15	64	164	40	40	164	64	15	5	576
Note 4. The remark about Mercury, that the cycles begin and end with the superior conjunction
according to the subtractive constant given, but the motions in the cycles begin and end with the
inferior conjunction, holds in the case of Venus also, showing thereby that the author is an ignorant
imposter, and cannot be VM. To correct the fault, 287¾ 'days' should be added or subtracted from
the subtractive constant.
Example 3. Compute Venus at 1,20,553 days from epoch.
If the subtractive constant given in the text is used, 1,18,817.5 (already found in the example in
Mars) – 1,18,122 = 695.5.
This × 2 ÷ 1151 leaves the remainder 240.
This divided by 2 gives 120 'days' gone in the cycle. We have for the first 5 days – 9°, and the next
15 days – 21° and the next 64 days – 15° and the remaining 36 days, 36 × 35 ÷ 164 = 7° 40', totally
– 37° 20'. Adding the Sun 17°.5 already found in the example for Mars, the true longitude of
Venus is – 20°, i.e. 340°. (The example in the Saurasiddhānta for the same date has given 46°.)
The error in Venus, in using this method here, is 66°. On the other hand, let us use the cycle
order re-arranged to begin from superior conjunction. It is 10° for 40 days, 35° for 164 days etc. We
have 10° for 40 days, and for the remaining 80 days, 80 × 35 ÷ 164 = 17°, total 27°. Adding the Sun,
17°.5, we have, true Venus, 44°.5. This is close to the correct 46°. This exposes the ignorance of the
imposter.
[शनिस्फुट:]
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