To finde the second inequality of Mars.
We must have given, 1 The angle N A S, which is to be found by subducting the ☉ place from the eccentrick of ♂ reduced, or this from it, so that lesse then 6 signes may remain, this remainer is the Anomaly of the Orbe, and the complement thereof is the angle N A S, or the halfe, is the halfe sum of the opposite angles.
- The eccentrick of ♂ reduced
- 257. 44576
- The ☉ true place
- 154. 07347
- Anomaly of the Orbe
- 103. 37229
- Complement is N A S
- 76. 62771
- Halfe Anomaly
- 51. 68614
As the greatest side N A | 146426 co. ar. | 4. 8343814 |
Is to Radius | 10. 0000000 | |
So is the lesser side S A | 100895 | 5. 0038707 |
To the tang. of | 34. 56887 | 9. 8382521 |
Adde | 45 | |
As Radius | ||
To co-tang. | 79. 56887 | 9. 2650444 |
So tang. ••/•• sum | 51. 68614 | 10. 1022929 |
To tang. ½ diff. | 13. 11536 | 9. 3673373 |
Summe | 64. 80150 angle A S N | |
Differ. | 38. 57078 angle A N S |
Because the Suns place was subtracted from the eccentrick of ♂ redu∣ced, therefore the angle of elongation A S N 64. 80150 must be added to the ☉ place 154. 07347 and then the place of ♂ will be 218. 87497.