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.