सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)
Siddhanta Shiromani Ganitadhyaya of Bhaskaracharya with Commentary
भास्कराचार्य द्वितीय द्वारा
पृष्ठ 423, कुल 573 में से
संदर्भ में पढ़ें408 body which lies between the position of the eclipsing body at the moment of first contact and the point of intersection with the path of the eclipsing body of the circle drawn with the centre of the eclipsed body as centre and radius equal to the difference of p+r -g where g is the magni- tude of the eclipse (grāsa) at the moment, or similarly the time taken by the centre of the eclipsing body to move through a similar and equal segment of the path of the eclipsing body on the other side, gives the time elapsed after the moment of first contact or the time before the moment of last contact. Comm. This is the converse of the above problem. The method is clear being based on rule of three as above. Both the problems could be algebraically expressed as follows. Let T, t, l, g, and k, stand respectively for the Sthiti-Khanda ie. the time between the moment of first contact to the middle of the eclipse or the time between the middle moment to the moment of last contact ; (2). the time elapsed after the moment of first contact or the time before the moment of last contact, as the case may be ; (3) the length of the Pragrahamārga or Mokṣamārga ; (4) the grāsa which is defined as p+r-k ; (5) the Karṇa whose expression is √(B²+β²), B being the Bhuja defined and β the latitude of the Moon. Then the following working is stipulated (a) If in time T, a path of length l is traced, what will be traced in t? The result is lt/T (b) Then B = l - lt/T (c) B² + β² = K² (d) p+r-k=g. Thus combining all the steps {l (T-t) / T}² + β² = (p+r-g)² ie. l² (T-t)² + β² T² = T² (p+r-g)² I given t, this equation gives g and given g it gives t.