Hour circle reading when } 11h. 57m. 57s. Telescope east. declination axis is horizontal. } 23h. 59m. 47s. Telescope west. ———— ———— ———— 12h. 1m. 50s. Error 0h. 1m. 50s. Or this error can be found and corrected without a level by taking two observations of a star of large declination in the same manner as in estimating the collimation error, for example:— Η URSÆ MAJORIS. Time by clock. Hour circle reading. 12h. 8m. 57s. 0h. 28m. 44s. Telescope east. 12h. 18m. 53s. 12h. 46m. 42s. Telescope west. ———— ———— ———— ———— ———— ———— 9m. 56s. 12h. 17m. 58s. 9m. 56s. ———— ———— 2) 8m. 2s. ———— ———— Error of hour circle due to error of 4m. 1s. inclination of axes[20] 6th Adjustment.—Bring the declination axis to a horizontal position with a level and set the hour circle to zero, or obtain the sidereal time from the nearest observatory, or again find it from the solar time by the tables, and correct it for the longitude of the place (subtracting the longitude reduced to time when the place is west and adding when east of the time-giving observatory) and set a clock or watch to it. Take the time of transit of a known star near the meridian and then the sidereal time by the clock at transit minus the right ascension of the star will give the hour angle past the meridian, and its difference from the circle reading is the index error, which is easily corrected by the vernier. If the star is east of the meridian the time must be subtracted from the right ascension to give the circle reading. In the above examples we have assumed, for the sake of better illustration, that the hour circle is divided into twenty-four hours, but more usually they are divided into two halves of twelve hours each. A movement through half a circle, therefore, brings the hour circle to the same reading again instead of producing a difference of twelve hours, as in the above example.