is there any formula for the perimeter (or length) of part of latitude A degree north of earth, which gets sun light, when tilted towards sun by an angle B degrees??
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$\begingroup$ It sounds like you are asking about the length of a circle of constant latitude. Of course if the oblate spheroid shape of the earth is taken into account, some measurements of the length of a degree of longitude (at given latitude) will provide an answer. Are you looking for a table of such data, or for a formula that fits the data? $\endgroup$– hardmathOct 8, 2014 at 15:10
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1 Answer
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If the Sun is at declination $\delta$, then at latitude $\beta$, the semi-diurnal arc $\Delta$ is given by $$ \cos(\Delta)=-\tan(\delta)\tan(\beta)\tag{1} $$ Assuming that the Earth is a sphere of radius $R$, we can adjust $(1)$ to get the length of the portion of a parallel of latitude along which the Sun is above the horizon, we get $$ 2R\cos(\beta)\arccos(-\tan(\delta)\tan(\beta))\tag{2} $$ The arccosine needs to be taken in radians.
For example, at mid-summer, the declination of the Sun is approximately $23^\circ27^\prime$ and the latitude of Bangalore is $12^\circ59^\prime$, assuming the radius of the Earth is $6371$ km, we get a length of $20747$ km.
Computation of Semi-Diurnal Arc
Since declination, latitude, and altitude are all measured from a point $90^\circ$ from the pole, the triangle of relevant angles on the sphere looks like
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where co-$\beta$ is the complement of the latitude, co-$\delta$ is the complement of the declination, $\Delta\lambda$ is the difference in hour angles or longitude, and co-alt is the complement of the altitude. The Spherical Law of Cosines says $$ \sin(\text{alt})=\sin(\delta)\sin(\beta)+\cos(\delta)\cos(\beta)\cos(\Delta\lambda)\tag{3} $$ When the Sun is on the horizon, it has an altitude of $0$. Thus, we get $$ \cos(\Delta\lambda)=-\tan(\delta)\tan(\beta)\tag{4} $$ which is the formula in $(1)$.
