ine of the latitude (=90deg) at the pole is unity, or 1; and the sine of
the latitude (=0deg) at the equator is 0. Therefore, at these two extremes,
the expression 15deg x sin. lat. actually does give the amount of _hourly
apparent rotation of the horizon_; namely, 15deg at one place, and 0deg at
the other. Now, as I understand the experiment, as given in the public
prints, it is asserted that the same expression of 15deg x sin. lat. will
give the _rotation of the horizon_ in intermediate latitudes; of which
rotation I subjoin a table calculated for the purpose.

+-----------+-------------+-----------------------+---------------------+
| | | Value of | Apparent |
| | Natural | 15deg x Sin. Lat., | corresponding |
| Degrees | Values of | or apparent | Times of _Horizon_, |
| of | Sine of the | _hourly_ Amount of | performing |
| Latitude. | Latitude. | Rotation of | one Rotation |
| | | _Horizon_, in Degrees | of 360deg, in Hours |
| | | and Decimals. | and Decimals. |
+-----------+-------------+-----------------------+---------------------+
| deg | | deg | h |
| 0 | 0.000 | 0.00 | Infinite time. |
| 1 | 0.017 | 0.26 | 1371.0 |
| 2 | 0.035 | 0.53 | 682.1 |
| 3 | 0.053 | 0.79 | 458.5 |
| 4 | 0.070 | 1.05 | 342.6 |
| 5 | 0.087 | 1.31 | 255.4 |
| 6 | 0.104 | 1.57 | 229.6 |
| 7 | 0.122 | 1.83 | 169.9 |
| 8 | 0.139 | 2.09 | 172.5 |
| 9 | 0.156 | 2.35 | 153.4 |
| 10 | 0.173 | 2.60 | 138.1 |
| 20 | 0.342 | 5.13 | 70.2 |
| 30 | 0.500 | 7.50 | 48.0 |
| 40 | 0.643 | 9.64 | 37.3 |
| 50 | 0.766 | 11.49 | 31.3