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[Phys-L] Re: Hours of sunlight



Re: Hours of sunlight
Regarding Marc "Zeke" Kossover's question:

Howdy-

Say I want to know the number of hours of sunlight
that a spot in my garden will get over a day. Say I am
not largely worried right now as to the total energy,
just the amount of sun.

If the spot is flat and there are no nearby walls, I
can figure the hours of light by looking up the
sunrise and sunset times and subtracting.

On the other hand, if there are nearby walls and
trees, I need to know the angle of the sun off the
horizon and the sun's path, the height of the wall and
distance to my patch of ground, and a bunch of trig to
figure out how much sun the area gets.

Does anyone know of a resource to help me make these
calculations?

Marc "Zeke" Kossover

Perhaps the following formulas will help.

Let's define the following angles:

L == local latitude (north being positive and south negative)

D == current declination of the sun from the celestial equator for
the current date (north being positive and south negative)

T == time *before* the local 'high noon', i.e. meridian transit,
measured as an *angle* of rotation of the earth (1 degree of
rotation corresponds to 4 minutes worth of time) This means a
positive T value is an AM time and a negative one is PM.

+ x direction == Horizontally East

+ y direction == Horizontally North

+ z direction == Vertically Up

Using these definitions the unit vector, u-hat = (u_x, u_y, u_z)
pointing *toward* the current location of the Sun in the sky is:

u_x = cos(D)*sin(T)
u_y = sin(D)*cos(L) - cos(D)*sin(L)*cos(T)
u_z = cos(D)*cos(L)*cos(T) + sin(D)*sin(L)

To find the hours of available sunlight first find the angle
the top of the wall blocking the early morning sun makes with the
true horizon (as measured from your fiducial spot on the garden's
surface). The sine of this angle is u_z when the sun is just
peeking up over that wall. Knowing D and L for your current date
and garden location use the above formula for u_z to calculate the T
angle for the morning appearance of the sun. Next find the angle the
top of the wall blocking the late afternoon sun makes with the true
horizon (again, as measured from your fiducial spot on the garden's
surface). Again set the sine of this angle to u_z and use the above
formula for u_z to find the T angle for the late afternoon
disappearance of the sun behind that wall. The difference between
these two T angles represents the time duration of the sunshine using
the conversion factor of 1 degree of angle equals 4 minutes of time.

Note: Since the above formula for u_z only involves the cosine of T
and since the cosine function does not distinguish positive from
negative angles, you will need to put the algebraic sign of T in by
hand. In particular, T is positive for the morning reading, and is
negative for the afternoon reading, and the difference between these
T-values is really the sum of their absolute values.

David Bowman
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