_To the Editor of the Scientific American_:
Given latitude N. 40° 51', declination N. 20° 25', sun 18° below the
horizon. To find the time of twilight at that place. In the
accompanying diagram, E Q = equinoctial, D D = parallel of
declination, Z S N a vertical circle, H O = the horizon, P = North
pole, Z = zenith, and S = the sun, 18° below the horizon, H O,
measured on a vertical circle. It is seen that we have here given us
the three sides of a spherical triangle, viz., the co-latitude 49° 9',
the co declination 69° 35', and the zenith distance 108°, with which
to compute the angle Z P S. This angle is found to be 139° 16' 5.6".
Dividing this by 15 we have 9 h. 16 m. 24.4 s., from noon to the
beginning or termination of twilight. Now, in the given latitude and
declination, the sun's center coincides with the horizon at sunset
(allowance being made for refraction), at 7 h. 18 m. 29.3 s. from
apparent noon. Then if we subtract 7 h. 18 m. 29.3 s. from 9 h. 16 m.
24.4 s., we shall have 1 h. 57 m. 55.1 s. as the duration of twilight.
But the real time of sunset must be computed when the sun has
descended about 50' below the horizon, at which point the sun's upper
limb coincides with the line, H O, of the horizon. This takes place 7
h. 16 m. 30.8 s. mean time. It is hoped the above will be a sufficient
answer to L.N. (See SCIENTIFIC AMERICAN of Dec. 1, 1883, p. 346.)