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