斜面の受ける日射量を求める簡単な一方法

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A very simple method was obtained for the computation of the intensity of the solar radiation on any slope. The principle is that the sun is fixed, and the slope revolves on the axis of the earth, and that the component of the normal to the slope to the solar radiation is equivalent to the intensity of the solar radiation on the slope.In Fig.1 OP is the normal to the slope, AB the cross line of the slope and horizontal plane.∠k=the angle of inclination, ∠h=azimuth.The components of OPare as follows;l=sin k sin h m=sin k cosh n=cos k}…(1) In Fig. 2 revole Y- and Z-axis on this X-axis for θ(=latitude) and make Y-axis parallel to the axis of the earth (Y') and then move the Y-axis on the axis of the earth. O'P' is the normal to the slope.Equations (1) are transformed as follows;l'=sin k sin h m'=sin k cos h cos θ+cos k sin θ n'=cos k cos θ-sin kcos h sin θ}…(2) Q' is the foot of the perpendicular from P' to Y'-axis.P'Q'={1-(sin k cos h cos θ+cos k sin θ)^2}^<1/2> Then revolve P'Q' on the Y"-axis (=the axis of the earth) with ω (=angular velocity of the earth).The motions of the components are expressed as follow;l"={1-(sin k cos h cos θ+cos k sin θ)^2}^<1/2>sin(ωt+α) m"=sin k cos h cos θ+cos k sin θ n"={1-(sin k cos h cos θ+cos k sin θ)^2}^<1/2>cos(ωt+α)}…(3) where α=tan^<-1>(sin k sin h)/(cos k cos θ-sin k cos h sin θ)…(4) Then in Fig. 4 turn the Y"-axis for δ (=declination). The component of the normal of the slope to the solar radiation is then obtained as follows;n'''={1-(sin k cos h cos θ+cos k sin θ)^2}^<1/2>cos(ωt+α) cos δ+(sin k cos h cos θ+cos k sin θ)sin δ…(5) Therefore, the intensity of the solar radiation on the slope is expressed as follows;I_s=I_0n'''=I_0[{1-(sink cos h cos θ+cos k sin θ)^2}^<1/2>cos(ωt+α)cos δ+(sin k cos h cos θ+cos k sin θ)sin δ]…(6) where I_0=intensity of the solar radiation.This is the equation which is used to calculate the intensity of the solar radiation on any slope.Also (6) can be integrated easily, and equation obtained (7).∫^<t_2>_<t_1>I_sdt=I_0[1/ω{1-(sin k cos h cos θ+cos k sin θ)^2}^<1/2>sin(ωt+α)cos δ+(sin k cos h cos θ+cos k sin θ)sin δ・t]^<t_2>_<t_1>…(7) where t_1,t_2 are the time of sunrise and sunset respectively.
一般社団法人日本森林学会の論文
1957-11-25

斜面の受ける日射量を求める簡単な一方法

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