Set the quadrant over both Stars at the observed degrees of altitude, and keep it fast upon the globe with your fingers; then slide the meridian up or down in the notches, ’till the quadrant cuts the given azimuth in the horizon; that elevation of the Pole will be the latitude required.
PROB. XLIII. _Having the Latitude of the place, to find the degree of the Ecliptic, which rises or sets with a given Star; and from thence to determine the time of its_ Cosmical _and_ Achronical _rising and setting._
Having rectified the globe for the latitude, bring the given Star to the Eastern side of the horizon, and mark what degree of the ecliptic rises with it: Look for that degree in the wooden horizon, and right against it, in the kalendar, you will find the month and day when the Star _rises Cosmically_. If you bring the Star to the Western side of the horizon, that degree of the ecliptic which rises at that time, will give the day of the month when the said Star _sets Cosmically_. So likewise against the degree which sets with the Star, you will find the day of the month of the _Achronical setting_; and if you bring it to the Eastern part of the horizon, that degree which sets at that time will be the Sun’s place when the Star _rises Achronically_.
Thus, in the latitude of _London_, _Syrius_, or the _Dog-Star_, rises _Cosmically_ the 10th of _August_, and sets _Cosmically_ the 10th of _October_. _Aldebaran_, or the _Bull’s Eye_, rises _Achronically_ on the 22d of _May_, and sets _Achronically_ on the 19th of _December_.
PROB. XLIV. _Having the Latitude of the place, to find the time when a Star rises and sets_ Heliacally.
Having rectified the globe for the latitude, bring the Star to the Eastern side of the horizon, and turn the quadrant round to the Western side, ’till it cuts the ecliptic in 12 degrees of altitude above the horizon, if the Star be of the first magnitude; then that point of the ecliptic which is cut by the quadrant, is 12 degrees high above the Western part of the horizon, when the Star rises; but at the same time the opposite point in the ecliptic is 12 degrees below the Eastern part of the horizon, which is the depression of a Star of the _first magnitude_, when she _rises Heliacally_; or has got so far from the Sun’s beams, that she may be seen in the morning before Sun-rising. Wherefore look for the said point of the ecliptic on the horizon, and right against it will be the day of the month when the Star _rises Heliacally_. To find the _Heliacal setting_, bring the Star to the West side of the horizon, and turn the quadrant about to the Eastern side, ’till the 12th degree of it above the horizon, cuts the ecliptic; then that degree of the ecliptic which is opposite to this point, is the Sun’s place when the Star _sets Heliacally_.