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

459
Fig. 109
it through the magnitude of ML the corresponding
Equatorial arc which is itself found through finding the
magnitude of AB. AB expressed in minutes will be equal
to the number of asus taken by the declination circle of G
namely PGB to have traversed from the position PDA to
PGB which is the time taken by the planet G to be at its
position from the moment of its rising at the Equatorial
horizon namely PDA, D being then its position. When
G was at D, K, the pole of the Ecliptic should have been
itself rising on the Equatorial horizon. Thus it is sought
to find the time expressed in asus taken by K from its
moment of rising on the Equatorial horizon to its present
position at K. When Bhāskara mentions in the commen-
tary under the verse that ‘‘क्षितिजकदम्बयोरन्तरं तदेवोत्तरमायनं
वलनम्’’ by the word क्षितिज, he had at the back of his mind
लङ्काक्षितिज ie. Equatorial horizon and not the horizon of
the place shown in the figure. Also as was mentioned
before in a previous context the time expressed in asus
taken by an equatorial arc to rise is equal to the number
of minutes in that arc. Since the time taken by the
planet to traverse the arc of the diurnal circle namely DG
is the same as the time taken by the equatorial arc LM
to rise, Bhāskara seeks to find the number of minutes in
AB, then compute, the number of Asus taken by LM to rise