Earth Eclipse of Server Sky Arrays
If the Earth was perfectly round, and the poles were not inclined, arrays in the 12789km, 17280 second radius equatorial orbit would spend 2868 seconds per orbit shaded by the 6371km radius Earth ( = 17280 \times asin( 6371 / 12789 ) / 180^o ).
In fact, the Earth has an equatorial radius of 6378.1 km, a polar radius of 6356.8 km, and an axial tilt of \phi = 23.439281° . The sun has an angular size of 0.53 degrees, and the Earth's atmosphere refracts light, meaning that the light dims gradually over approximately 30 seconds entering eclipse. For the rest of this analysis, we will ignore these gradual effects, pretend the sun is a point source at infinity, and calculate the hard cutoff time as a function of time of year.
The variable \beta represents the time of year in the northern hemisphere, from 0° in spring, 90° in summer, 180° in the fall, and 270° in winter.
Simple Round Earth
From the viewpoint of the sun, at the beginning of spring at the vernal equinox, the m288 orbit is in a plane tilted \phi = 23.439281° clockwise, to the right. The equations for this orbit are:
x_0 = R_{288} \sin( \phi ) \cos( \theta ) ~ ~ ~ , ~ ~ ~ y_0 = R_{288} \sin( \theta ) ~ ~ ~ and ~ ~ ~ z_0 = R_{288} \cos( \phi ) \cos( \theta )
Over the course of a year, this orbit is rotated clockwise around the z axis by the angle \beta .
x = x_0 \cos( \beta ) + y_0 \sin( \beta ) ~ ~ ~ , ~ ~ ~ y = -x_0 \sin( \beta ) + x_0 \cos( \beta ) ~ ~ ~ and ~ ~ ~ z= z_0
x = R_{288} \bigl( \sin( \phi ) \cos( \theta ) \cos( \beta ) + \sin( \theta ) \sin( \beta ) \bigr)
y = R_{288} \bigl( - \sin( \phi ) \cos( \theta ) \sin( \beta ) + \sin( \theta ) \cos( \beta ) \bigr)
z = R_{288} \bigl( \cos( \phi ) \cos( \theta ) \bigr)
The orbit passes into and out of shadow where y > 0 ( to the back ) and x and z are on a projected circle with radius R_E = 6371 km or x^2 + z^2 = R_E^2 .
\bigl( { {R_E} \over {R_{288}} } \bigr)^2 = \bigl( { \sin( \phi ) \cos( \theta ) \cos( \beta ) + \sin( \theta ) \sin( \beta ) } \bigr)^2 + \bigl( { \cos( \phi ) \cos( \theta ) } \bigr)^2
This equation can be solved analytically for cos( \theta ) , but the result is very messy! We can assign some variables along the way.
k \equiv \bigl( { {R_E} \over {R_{288}} } \bigr)
p \equiv \sin( \phi ) \cos( \beta )
q \equiv \sin( \beta )
r \equiv \cos( \phi )
c \equiv \cos( \theta )
s \equiv \sin( \theta ) ~ ~ ~ ~ note: ~ 1 = c^2 + s^2
The equation becomes:
k^2 = ( p c + q s )^2 + ( r c )^2
After some juggling we get:
( k^2 - q^2 ) = ( p^2 + r^2 - q^2 ) c^2 + 2 p q c s
Assign some more variables:
d \equiv k^2 - q^2
e \equiv p^2 + r^2 - q^2
f \equiv p q
So
d = e c^2 + 2 f c s
( d - e c^2 )^2 = 4 f^2 c^2 s^2 = 4 f^2 c^2 - 4 f^2 c^4
Promising! We have an equation that is quadratic c2. We can massage this into the quadratic equation:
c^2 = { { { 2 f^2 + d e } \pm { 2 f \sqrt { f^2 + d e - d^2 } } } \over { 4 f^2 + e^2 } }
Compute \theta = \pm \arccos( \pm c ) , and get 8 values for \theta , some complex. By choosing the two real values where y > 0 , and differencing them, we compute the angle spent in shadow. Here is a graph of the result:
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Oblate Earth
From a distant viewpoint in the equatorial plane, the earth can be approximated as an elliptical disk, with a semimajor X axis of 6378.1 kilometers, and a semiminor Z axis of 6356.8 km. The edge of this disk follows the equation:
1 ~ = ~ \bigl( { {x_e} \over { 6378.1 km } } \bigr)^2 + \bigl( { {z_e} \over { 6356.8 km } } \bigr)^2
But the Sun is not on the equatorial plane, so from the Sun the Earth looks tilted, with the north end towards the front during northern summer, back in northern winter, to the left during northern fall, and to the right during northern spring. then the projection of the oblate spheroid onto the XZ plane is MORE LATER