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

526 Fig. 125 Refer to fig. 125. Gr NK is the celestial equator. RrL is the ecliptic and RGAM the Moon's orbit where R is the Rāhu or ascending node of the lunar orbit. Let the obliquity of the ecliptic be ω (omega) and the inclination of the lunar orbit to the ecliptic be i. ω was taken to be 24° and i 4½° by Bhāskara. Let the lunar orbit RGAM cut the celestial equator in G which is called the Moon's Gola Sandhi. r is the Sun's Gola Sandhi. Bhāskara first wants us to locate G ie. to find rQ which gives its longitude. Let A be the position of the Moon when his celestial longitude is zero ie rA is perpendicular to the ecliptic. Let M be the position of the Moon when his celestial longitude is 15° (Here the figure is not drawn to scale but a little exaggerated for the sake of clarity). Let ML be the celestial latitude of the Moon in its position. M. If LK be drawn perpendicular to the equator, LK is called the Asphuṭa - krānti of the Moon. ML is called the Asphuṭa-Vikṣepa of the Moon. MN drawn perpendicular to the celestial equator ie. the declination of the Moon is called Sphuṭakrānti of the Moon. Draw perpendicular LS on MN. Then MS is called the Sphuṭa Vikṣepa of the Moon. Since MN = Ms + LK Sphuṭakrānti = sphuṭa Vikṣepa + Asphuṭakrānti. Let Ap be the declination of the Moon when his longitude is zero.