On Fri, 20 Mar 1998, reto ambrosini <[EMAIL PROTECTED]>
wrote:
>I tried to calculate the sunrise time with the formula
>
>cos(T)=-tan(lat)*tan(decl)
>
>I took the datas from the astronomical Almanac, here is an
>example for the 3. of september 1998 at latitude +46
>
>Sun declination : 7 42 13.5 = 7.704 deg
>Rise time : 5h23m
>Set time : 18h35m
>Day length : 13h12m
>
>T=arccos(-tan(46)*tan(7.704))=98.05 deg =6.537 hours
>
>Calculated rise time : 12-6.537 = 5.463h = 5h27m47s
>Calculated set time : 12+6.537= 18.54h = 18h32m13s
>Calculated day length : 6.537*2 = 13.07 h = 13h04m25s
Dear Reto,
There are several reasons for the discrepancy:
(1) The sun's declination changes, even in the course of one day. So, after
getting an approximate time of sunrise or sunset, you should really find the
sun's declination *at those times* and repeat the calculation.
(2) The Astronomical Almanac defines sunrise as the time when the sun's
calculated altitude is 90d -50', not 90d 00'. This allows 34' for refraction
and 16' for the sun's semidiameter, and the result is supposed to be the time
when the uppermost tip of the sun's limb is just visible on the horizon. So
your formula should be modified as follows:
cos(T) = [cos(90d 50')-sin(lat)*sin(decl)]/[cos(lat)*cos(decl)]
(3) The semidiurnal arc, T, when doubled, gives the length of a day in
*sidereal* time. Divide this interval by 1.002737 to convert to solar time.
(4) If you need the local mean time of sunrise or sunset, it is necessary to
take account of the equation of time (calculated by the methods discussed on
this list in recent months). That is, the sun often transits the local
meridian
a few minutes before or after 12 noon, rather than at noon exactly.
Sincerely yours,
Roger Sinnott
Sky & Telescope
