Daylength by latitude
Margaret Baker
latitude?
I know that each location has sunrise/sunset tables, but we are looking
for a way to approximate daylength based on latitude .... Any
suggestions on who to ask or where to find the info?
Thanks, Marg
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Charles & Margaret Baker, Portland, Oregon, USA (cmb...@teleport.com)
"Orchid Species Culture" - Timber Press. Vol.1, Pescatorea, Phaius,
Phalaenopsis, Pholidota, Phragmipedium, Pleione - ISBN 0-88192-208-0
Vol. 2, in press, Dendrobium
Bear Giles
>Can anyone point us to a source for determining daylength based on
>latitude?
>I know that each location has sunrise/sunset tables, but we are looking
>for a way to approximate daylength based on latitude .... Any
>suggestions on who to ask or where to find the info?
If you mean that literally, i.e., if you want *only* the length of the
day, you can get a decent first approximation using straightforward
trigonometry.
1. Draw a circle.
2. Draw a line at the latitude in question.
2. Draw a "terminator" line at 23.5 * sin (day-of-year/365) degrees,
to fudge for the apparent movement of the sun throughout the year.
N.B. I'm starting the year at an equinox; you'll need a "phase angle"
in the trig term.
4. Measure what fraction of the line is on the "daylit" side of the
line.
5. Using simple trig, figure out what fraction of the *circle* the
sunlight part of the line corresponds to. That fraction is identical
to the fraction of the day in (direct) sunlight.
This approximation has several limitations:
1. You don't have any information on when sunrise/sunset occurs. Since
the earth's orbit is not perfectly circular, sunrise/sunset is not
symmetrical around the average "local noon". BTW, the "figure 8"
on some globes deals with this.
2. Your terminator line may be slightly off, once again since the
earth's orbit is not perfectly circular. This is a minor factor.
3. This approximation does not take into consideration atmospheric
effects, esp. twilight. Extending the terminator line a half-degree
or so towards the dark side will give a decent approximation for
the atmospheric refraction extending sunrise/sunset, and a couple
degrees will give you an approximation for twilight. This
approximation will not work well at latitudes near to the point
where the sun is just up (or just down) all 24 hours.
Many years ago I wrote a program based on this approach. At 40 N, it
was generally accurate on the day length to within a few minutes
(when it included the half-degree approximation for sunrise/sunset),
but was often 10-15 minutes off on the actual sunrise time. On the
brighter side, the error on sunrise times was pretty consistent -- the
error from day-to-day was pretty close.
--
Bear Giles
be...@fsl.noaa.gov
Bear Giles
(This applies to the Northern Hemisphere, btw...)
Everyone knows that the shortest day of the year is winter equinox,
but relatively few people know that the day with the earliest sunset
time is a few days *before* the equinox.
This surprising fact is due to the earth passing perihelion a few
weeks before the equinox. So while the days are getting shorter,
the "solar noon" is also getting later due to orbital motion.
Very close to solstices the length of the day is nearly uniform,
but since "solar noon" is a few minutes later each day sunset
occurs later and later, and hence the earliest sunset is a few
days before the winter solstice.
I haven't checked, but I think the corresponding effect near the
summer solstice is much less pronounced and may be masked by the
solstice itself. This is because the earth has just passed apohelion
and is moving relatively slowly.
--
Bear Giles
be...@fsl.noaa.gov
Shawn Trueman
Margaret Baker <cmb...@teleport.com> wrote:
>Can anyone point us to a source for determining daylength based on
>latitude?
>I know that each location has sunrise/sunset tables, but we are looking
>for a way to approximate daylength based on latitude .... Any
>suggestions on who to ask or where to find the info?
>
>Thanks, Marg
>
>----------------------------------------------------------------------
>
the astronomical computing section called "Sunshine."
--Shawn Trueman
jaba...@delphi.com
A reasonably accurate estimation of day length can be obtained using
2 simple equations; one to calculate solar declination angle, and another
to calculate day length. Neglecting parallax and atmospheric refraction,
a geometric relationship (which can be seen from spherical triginometry)
relating solar zenith and hour angles is:
cos(Z) = sin(L) sin(D) + cos(L) cos(D) cos(H) , [1]
in which Z is solar zenith angle (angle to sun from sky zenith), H is
solar hour angle (angle in deg longitude from the subsolar point to a
location), L is latitude, and D is solar declination angle. All angles
are expressed in deg. At sunrise and sunset, solar zenith angle is
approximately 90 deg. Due to atmospheric refraction and the sun's
finite dimensions, it is slightly greater - I believe about 90.8 deg.
At Z = 90 deg, cos(Z) = 0, such that solar hour angle at sunrise and/or
sunset can be estimated from
cos(H) = tan(L) tan(D) . [2]
Since a solar hour angle of 15 deg corresponds to 1 hr of time, and day
length is twice that from sunrise (or sunset) to solar noon, an equation
for day length (Q) in hr is
Q = 2/15 arccos(tan(L) tan(-D)) . [3]
The negative of solar declination angle is used to compute the arrcos in
the proper quadrant - there are 2 values of H which satisfy this
equation (equal in magnitude, but of opposing sign).
Solar declination can be calculated as Bear pointed out, using
D = 23.44 sin(360 (J - 82) / 365.25) , [4]
in which J is Julian calendar day. This is slightly different from his
equation, and gives more accuracy over the next 12 years. Athough being
quite accurate near Summer solstice (when solar energy is most abundant
in the Northern Hemisphere), it is off by almost 1 deg in late fall and
late winter. Those errors result in ones for Q of generally < 6 min for
low and middle latitudes. For high latitudes (> 50 deg) however, errors
can be greater than 15 min near Winter solstice. A more accurate
expression for solar declination is
D = .33029 - 22.9717 cos(X) + 3.8346 sin(X) - .3495 cos(2X) + [5]
.0261 sin(2X) -.1392 cos(3X) + .0727 sin(3X) ,
in which X = J(360/366). This is quite accurate near the solstices,
and has a maximum calculation error of .16 deg near the equinoxes. The
limiting error in [5] is leap days, since D can change by .4 deg during
an entire day, approaching or just past an equinox.
As mentioned, parallax and especially atmospheric refraction cause
day length to be different than that calculated above. Parallax is
very small, and shortens days. Refraction is much larger, and lengthens
days; typically by only 5-10 min at low and middle latitudes. This can
be much greater at high latitudes, especially during very long days.
Below is a table showing day length (hr:min) at various Northern
Hemisphere latitudes at Summer solstice. Also included is increase in
day length (min) due to factors neglected in the calculation method
shown above.
Latitude Day length Increase
0 12:06 5
20 13:19 6
30 14:03 7
40 14:58 8
50 16:19 10
55 17:18 12
59 18:25 16
62 19:38 21
64 20:51 29
65 21:47 39
65.5 22:29 52
65.7 22:53 62
65.8 23:08 70
65.9 23:34 87
~65.94 near 24:00 > 100
One should keep 2 things in mind regarding this: 1) these are for Summer
solstice, and can be somewhat different at other times of year, and 2)
the rapid increase in time is not very meaningful, since in most places
the sun cannot even be seen in these circumstances. I am not certain
about the refraction equation - if anyone sees that these numbers are
significantly wrong, please let me know.
In case you are also interested in the time of solar noon (N) in hr
standard time, it can be calculated as follows:
N = 12 + (P - Pr)/15 - E , [6]
in which P is longitude, Pr is reference longtiude for standard time,
and E is magnitude of the "equation of time". In a similar manner to
[5], E can be calculated (in hr) as
E = -.0001062 + .009554 cos(X) - .122671 sin(X) - [7]
.053350 cos(2X) - .156121 sin(2X) .
This equation is most accurate at times of high solar declination angle
also, with a maximum calculation error of only 29 sec; which is
approximately its maximum daily change.
Perhaps this is more info than you were looking for. However, I
think it will be very helpful. More accuracy can be obatined using
more (complicated) equations; but using equations [4] or [5] and [3],
with an appropriate correction for refraction (typically 5-10 min), day
length can be calculated accurate to within a few min. Use of [6] and
[7] allows one to calculate sunrise and sunset times just as accurately.
Joseph Bartlo
jaba...@delphi.com
to save someone the trouble of doing this for me. I knew that minus sign
belonged somewhere........Joseph