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### Daylight Hours and Latitude

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Date: 9 Jan 1995 11:50:06 -0500
From: Chianne
Subject: Question

Hello, I'm a student at Monta Vista High School, in Cupertino CA. I am
currently enrolled in Calculus and my teacher had a question that I was

How does the number of daylight hours on a given day of the year depend
on latitude.  State dependents mathematically.

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Date: 13 Jan 1995 02:23:47 -0500
From: Dr. Ken
Subject: Re: Question

Hello there!

You know, my first instinct is to say that the number of daylight hours in a
given _year_ isn't dependent at all on latitude.  Let's build up a model for
the earth's rotation and revolution, and see whether this works out.  See,
my idea is that the hours you lose when you're tilting away from the sun are
made up in full when you're tilting toward the sun.  But let's make sure.

Okay, well, first let's pretend the earth has no tilt, that its axis of
rotation is perpendicular to its plane of revolution.  Then every point on
the earth (except the north and south poles) will have the same pattern
every day: half the day is sunny, and half the day is dark, so in a year the
daylight percentage will be 50%.  So that case is killed.  So it looks like
rotating won't all by itself skew daylight hours.

But the earth isn't made like that.  Too bad.  But I still think this has a
lot of applicability.  See, let's now pretend that instead of there being no
tilt, there's no rotation, and the earth just flies around the sun without
rotating on its own axis.  Then every part on the earth is sunny for half
the year, and dark for the other half, and the daylight percentage is again
50%.  So revolution won't by itself skew daylight hours, either.

So is it reasonable to guess that the two combined won't skew daylight hours
too?  It seems like it to me.  I'll see if I can get you started on showing
why.

Let's pick a point on the earth without loss of generality, assume it's in
the northern hemisphere, because the same exact stuff is going to happen in
the southern) and let l be the latitude of our point (for calculation's
sake, let l be in the range 0 to 90, where l=0 means that our point is the
north pole and l=90 means that we're on the equator).  Then let's press the
start button on the earth at sun-up on the summer solstice, when the top of
the earth is tilting its most toward the sun.  Now let's introduce a dummy
variable, v, the tilt of the earth relative to the sun (v will go from
something like 20 degrees to -20 degrees, and then end up at 20 again - I'm
hoping that the actual measure of the tilt of the earth won't really make
any difference).

So if the tilt of the earth on a certain day is l degree, what portion of
the day is blessed with daylight?  Well, there are a couple of cases to
consider here: it could be that our point is above the arctic circle (or
below the antarctic circle) and that the whole day is either all sun or all
non-sun.  But we don't really have to consider this very much, since for all
these points there will be a corresponding time six months later that will
even things out: if it was sunny on that point, it will be shady for the
same amount of time, and vice versa.  This just has to do with the tilt of
the earth, and not the rotation: since the whole day is sunny or shady,
spinning the earth isn't going to do anything to the number of daylight
hours at that point.

Do you believe that?  If so, keep reading.  If not, see if you can puzzle
through it.

Anyway, in figuring out what portion of our day is sunny at our chosen
point, we can assume that our point gets some sun and some shade, i.e.
that Abs[v] (is less than) l (is less than) (Abs[v] means the absolute value
of v).  So I'll leave it to you to determine what portion of the day is sunny,
and you'll get an answer in terms of v and the constant l.

So then you'll have to figure out what happens when you let v go from
20 to -20 to 20.  Will things cancel out of the final daylight sum, or fit
together nicely or anything?  I bet they will.

assumed that the tilt of the earth stays constant for a whole day, then
changes all at once for the next day.  In reality, the tilt is changing
continuously, of course.  But you shouldn't be too far off doing it this
way.  It's kind of like using a Riemann sum to approximate an integral.

Secondly, I've kind of ignored the "Mercury Effect."  I read somewhere
(when I was about five, so I could be way off on this) that Mercury
revolves at the same rate it rotates, so that the same side of the planet
always faces the sun.  So the number of daylight hours doesn't depend
on the true latitude, it depends on the relationship to a different axis
altogether, the axis which is perpendicular to the plane of revolution.

Well, we don't have quite the same problem on earth, since we have our
days so much faster than our years.  But the same kind of thing happens,
a little bit, because our years and days don't end up finishing exactly at
the same time.  But since that could make it really hairy and wouldn't
change the answer much, I decided to ignore it.

Anyway, this is an interesting problem in modeling.  I hope you enjoy
working with it!

-Ken "Dr." Math
http://mathforum.org/dr.math/

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Date: 12/09/2001 at 18:17:47
From: James
Subject: Re: Daylight Hours and Latitude

Though I enjoyed your observations, I believe you have left out
something vital.

Because the earth's orbit is an ellipsoid rather than a circle, the
earth spends more time with the northern hemisphere tilted towards
the sun, and less time with the northern hemisphere tilted away from
the sun.

I base this statement on the fact that the earth slows down and
travels farther during that part of the ellipsoid that its distance
from the sun is greater, and speeds up and travels a shorter distance
during that part of the ellipsoid that it is closer to the sun.

From the dictionary's identification of the approximate occurrence of
the vernal equinox near March 21, and the autumn equinox near
September 23, that time difference isn't very large, but it is real.

Moreover, its effect will be to increase the number of the northern
hemisphere's longer summer days the most and shorten the number of
the shortened winter days the most, at the solstices.  (Sorry, that's
the best I can do.  The underlying point is that the earth spends a
longer period of time at the northern hemisphere's summer end of its
orbit then it does at the winter end of its orbit - hence a shift to
more time of more sunlight during the summer and less time of less
sunlight during the winter. The effects are not symmetrical.)

I also recall a lopsided figure eight on some globes of the earth,
which I believe depicted the sun's apparant latitude throughout the
year.

I would conclude that the northern hemisphere gets more sunlight, and
the increase is more as you move to the north pole.  The southern
hemisphere gets less sunlight, and the imbalance between sunlight and
darkness increases as you go farther south.

James

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Date: 12/09/2001 at 21:06:53
From: Doctor Rick
Subject: Re: Daylight Hours and Latitude

Hi, James.

Your idea sounds reasonable to me. The perihelion (point when the earth
is closest to the sun) occurs in northern-hemisphere winter. (I remember
this because it is counter to the naive notion that it's cold in winter
because the sun is farther away.) The earth moves faster along its orbit
when it is closer to the sun (due to Kepler's law that a body sweeps out
equal areas in equal times). Thus the earth spends less time with shorter
days in the northern hemisphere than with longer days.

You might try to confirm your qualitative hypothesis using a formula that
a correspondent sent in to us. See this page in our Dr. Math Archives:

Latitude and Longitude and Daylight Hours
http://mathforum.org/dr.math/problems/fiva9.21.98.html

I think you'll find the page interesting. After the initial exchange, you
will see the formula passed to us by David Toomey. You could use this
formula (most easily, perhaps, in a spreadsheet) to calculate the total
number of daylight hours in a year at various latitudes. I'd like to hear
back from you if you follow up on this!

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/

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Date: 09/23/2002 at 13:04:54
From: Jim
Subject: Re:  Daylight Hours and Latitude

There is an additional factor that has been omitted so far:  refraction.
A person can see the sun after geometry would say the sun is below the
horizon because the atmosphere bends the light.  This is especially
obvious today, the autumnal equinox, where every place on Earth gets
_more_ than 12 hours of sunlight.  Check sunrisesunset.com, for example,
and you'll see this.  Here are a few geographically diverse examples:

Anchorage,                    12:15
Stanley, Falkland Islands,    12:10
Honolulu,                     12:08
Bogota,                       12:06
Caracas,                      12:07

High latitudes (Northern and Southern Hemisphere alike) get more sun
than equatorial sites today.  Balancing this, perhaps, is that
refraction won't make the day any longer at the North Pole in the
summer because the North Pole already gets 24 hours of sun.

-Jim

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Date: 03/20/2007 at 14:06:31
From: Stephen
Subject: Daylight Hrs & Lat - Refraction effect more at pole - why?

The last posting discussing refraction and latitude on the equinox
shows that the further north you go the longer day you have.

What causes this?  Is it that the earth bulges out around the equator
and the poles are flatter?  That the curvature of the earth at the
equator is greater and we pass through the refracted light quicker
than at the pole where the curvature of the earth is less and we pass
through the refracted light slower?

Or is it that the earth is still a squished ball but the refractive
effects of the atmosphere cause greater refraction at the poles than
the equator?

Thanks for you help!

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Date: 03/20/2007 at 17:55:11
From: Doctor Rick
Subject: Re: Daylight Hrs & Lat - Refraction effect more at pole - why?

Hi, Stephen.

I can think of one cause that might account for this small effect.
At the equator the sun drops straight down to the horizon at the
equinoxes.  At high latitudes such as that of Anchorage or Reykjavik,
the sun sets at a relatively shallow angle to the horizon.  Thus it
takes longer at higher latitudes for the sun to get a given angular
distance below the horizon (based on straight lines).  If refraction
causes the visible sunset to occur when the sun is actually a
certain distance below the horizon, this will occur later at high
latitudes.  (Likewise the visible sunrise will occur earlier there.)

By the way, the refraction issue is mentioned in the archive page to
which I referred in the next-to-last message.  Also mentioned there
in passing is the fact that sunset is reckoned as when the TOP of
the sun goes below the horizon (or appears to), not the CENTER of
the sun.  This in itself lengthens the day at the equinoxes, and
again this effect is greater where the sun's path makes a smaller
angle to the horizon.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/

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Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Physics/Chemistry
High School Practical Geometry

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