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A solution for sunrise and sunset hour angles on a tilted surface without a singularity at zero azimuth
Solar Energy Vol. 28, No. 4, p. 357, 1982 Printed in Great Britain.
O038-092XlS2/040357--Ol$03.~lO Pergamon Press Ltd.
TECHNICAL NOTE A solution for sunrise and sunset hour angles on a tilted surface without a singularity at zero azimuth ROBERT H . BUSHNELL 502 Ord Drive, Boulder, CO 80303, U.S.A. (Received 22 October 1980; revision accepted 23 April 1981) The letter to the editor by Andersen [1], about possible confusion of east and west in calculating sunrise and sunset hour angles, repeats the expressions for these angles given by Klein[2]. A derivation can be found on p. 306 of Jones [3]. These contain the tangents of both tilt and azimuth in denominators. Consequently there is a singularity at zero flit and zero azimuth. This leads to the use of different expressions for east and west with the resulting need to keep track carefully of sign. This is not necessary. The following development gives an expression without the singularity. Entering the sign correctly is all that is needed. Tests for east and west are not required. The angle between the sun and the normal to a surface is implied in (1), taken either from Andersen[i] or (2.5.2) on p. 15 of Duttie and Beckman[4], where the terms in cos ~o and sin o~ have been collected. Angles for a tilted surface are shown in Fig. 1. Azimuth is positive clockwise starting from true .south. cos 0
=
sin 8(sin ~ cos s - c o s ~ sin s c o s y ) + cos 8(cos ¢ cos s + sin ¢ sin s cos y)cos co
By trigonometric identity, this equals cos(~oo-x) or ~ o - x = arccos(- a/D).
(6)
Now we must replace x by its definition from (4). Here again there is a choice. If we use the cosine we get an ambiguous solution, requiring that every time o~o is calculated we have to decide which sign to use. We use the sine. ~Oo= arccos(- a/D) + arcsin(c/D). (7) Only the positive square root is used. The denominator does not go to zero so singularities do not occur. Notice that c contains the sine of the azimuth angle so the solution goes smoothly through zero. There are two angles in the range - ¢r to + ¢r which produce a given cosine in (7). One is for sunrise. The other is for sunset. Because most computers and hand calculators return the positive angle, we can write
(1) co, = - arccos( - a/D) + arcsin(c/D)
(8)
~o, = arccos( - a/D) + arcsin(dD)
(9)
+ cos 6 sin s sin y sin co. Now let (2)
cos O = a + b cos to + c sin to
where a, b and c are the corresponding factors in (1). We take the declination 8 to be constant. The values a, b and c are constant as the sun traverses the sky. To find the sunrise and sunset hour angles ~, and ~o, we set cos 0 equal to zero. Having done this in (2) and having transposed, we divide by (b 2 + c2)tt2 which, for convenience in printing, we call D. (3)
- aiD = (b cos O~o)/D+ (c sin tOo)lD
where wo is the hour angle which makes cos 0 = 0. Now we make a trigonometric substitution. There is more than one choice for this. Some choices do not give both sunrise and sunset. The best choice is
(4)
biD = cos x and c/D = sin x which gives
(5)
- aiD = cos x cos ~Oo+sin x sin OJo.
where a, b, c and therefore D are defined in (1) and (2). If either aid or c/D is outside the range - 1 to + 1 no sunrise occurs. For s - - 0 (8) and (9) reduce to the usual arceos( - tan 8 tan ~,). These relations are correct in both the northern and southern hemispheres. The angle definitions are taken for convenient use in the northern hemisphere. In the southern hemisphere the azimuth angle is still taken positive clockwise (positive west) from true south so the azimuth is usually out the back of the surface. The tilt angle is positive up toward the direction of the azimuth so a tilt toward the equator is usually negative in the southern hemisphere. NOMENCLATURE
y azimuth angle, positive clockwise, zero at true south 8 declination of the sun co hour angle of the sun wo hour angle of the sun when cos 0 is zero oJ, sunrise hour angle w, sunset hour angle latitude 0 angle between the direction to the sun and the normal to the surface s tilt angle up from horizontal toward the direction of the azimuth
R F - ~ I ~ N C F.,S s
uth
Fig. 1.
1. P. Andersen, Comments on "Calculation of monthly average insolation on tilted surfaces" by S. A. Klein. Solar Energy 25, 287 (1980). 2. S. A. Klein, Calculation of monthly average insolation on tilted surfaces. Solar Energy 19, 325 (1977). 3. R. E. Jones, Effects of overhang shading of windows having arbitrary azimuth. Solar Energy 24, 305 (1980). 4. J. A. Dufile and W. A. Beckman, Solar Energy Thermal Processes. Wiley-Interscience, New York (1974). 357
O038-092XlS2/040357--Ol$03.~lO Pergamon Press Ltd.
TECHNICAL NOTE A solution for sunrise and sunset hour angles on a tilted surface without a singularity at zero azimuth ROBERT H . BUSHNELL 502 Ord Drive, Boulder, CO 80303, U.S.A. (Received 22 October 1980; revision accepted 23 April 1981) The letter to the editor by Andersen [1], about possible confusion of east and west in calculating sunrise and sunset hour angles, repeats the expressions for these angles given by Klein[2]. A derivation can be found on p. 306 of Jones [3]. These contain the tangents of both tilt and azimuth in denominators. Consequently there is a singularity at zero flit and zero azimuth. This leads to the use of different expressions for east and west with the resulting need to keep track carefully of sign. This is not necessary. The following development gives an expression without the singularity. Entering the sign correctly is all that is needed. Tests for east and west are not required. The angle between the sun and the normal to a surface is implied in (1), taken either from Andersen[i] or (2.5.2) on p. 15 of Duttie and Beckman[4], where the terms in cos ~o and sin o~ have been collected. Angles for a tilted surface are shown in Fig. 1. Azimuth is positive clockwise starting from true .south. cos 0
=
sin 8(sin ~ cos s - c o s ~ sin s c o s y ) + cos 8(cos ¢ cos s + sin ¢ sin s cos y)cos co
By trigonometric identity, this equals cos(~oo-x) or ~ o - x = arccos(- a/D).
(6)
Now we must replace x by its definition from (4). Here again there is a choice. If we use the cosine we get an ambiguous solution, requiring that every time o~o is calculated we have to decide which sign to use. We use the sine. ~Oo= arccos(- a/D) + arcsin(c/D). (7) Only the positive square root is used. The denominator does not go to zero so singularities do not occur. Notice that c contains the sine of the azimuth angle so the solution goes smoothly through zero. There are two angles in the range - ¢r to + ¢r which produce a given cosine in (7). One is for sunrise. The other is for sunset. Because most computers and hand calculators return the positive angle, we can write
(1) co, = - arccos( - a/D) + arcsin(c/D)
(8)
~o, = arccos( - a/D) + arcsin(dD)
(9)
+ cos 6 sin s sin y sin co. Now let (2)
cos O = a + b cos to + c sin to
where a, b and c are the corresponding factors in (1). We take the declination 8 to be constant. The values a, b and c are constant as the sun traverses the sky. To find the sunrise and sunset hour angles ~, and ~o, we set cos 0 equal to zero. Having done this in (2) and having transposed, we divide by (b 2 + c2)tt2 which, for convenience in printing, we call D. (3)
- aiD = (b cos O~o)/D+ (c sin tOo)lD
where wo is the hour angle which makes cos 0 = 0. Now we make a trigonometric substitution. There is more than one choice for this. Some choices do not give both sunrise and sunset. The best choice is
(4)
biD = cos x and c/D = sin x which gives
(5)
- aiD = cos x cos ~Oo+sin x sin OJo.
where a, b, c and therefore D are defined in (1) and (2). If either aid or c/D is outside the range - 1 to + 1 no sunrise occurs. For s - - 0 (8) and (9) reduce to the usual arceos( - tan 8 tan ~,). These relations are correct in both the northern and southern hemispheres. The angle definitions are taken for convenient use in the northern hemisphere. In the southern hemisphere the azimuth angle is still taken positive clockwise (positive west) from true south so the azimuth is usually out the back of the surface. The tilt angle is positive up toward the direction of the azimuth so a tilt toward the equator is usually negative in the southern hemisphere. NOMENCLATURE
y azimuth angle, positive clockwise, zero at true south 8 declination of the sun co hour angle of the sun wo hour angle of the sun when cos 0 is zero oJ, sunrise hour angle w, sunset hour angle latitude 0 angle between the direction to the sun and the normal to the surface s tilt angle up from horizontal toward the direction of the azimuth
R F - ~ I ~ N C F.,S s
uth
Fig. 1.
1. P. Andersen, Comments on "Calculation of monthly average insolation on tilted surfaces" by S. A. Klein. Solar Energy 25, 287 (1980). 2. S. A. Klein, Calculation of monthly average insolation on tilted surfaces. Solar Energy 19, 325 (1977). 3. R. E. Jones, Effects of overhang shading of windows having arbitrary azimuth. Solar Energy 24, 305 (1980). 4. J. A. Dufile and W. A. Beckman, Solar Energy Thermal Processes. Wiley-Interscience, New York (1974). 357