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| 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 $ ). | 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^\circ ~ $ ). |
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| == 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! Best to let Mr. Computer take care of it: MORE LATER |
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| 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: | The equatorial plane is tilted towards the sun by angle $ \theta_{eq} $ defined by: |
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| $ 1 ~ = ~ \bigl( { {x_e} \over { 6378.1 km } } \bigr)^2 + \bigl( { {z_e} \over { 6356.8 km } } \bigr)^2 $ | $ \sin( \theta_{eq} ) = \sin( \beta ) \sin( \phi ) ~ ~ ~ $ see [Precession] |
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| 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 |
The earth can be approximated as an elliptical disk, a projection of a ellipsoidal spheroid with an equatorial radius $ R_E $ = 6,378,137 meters and a polar radius $R_P$ = 6,356,752 meters. The edge of this elliptical disk follows the equation: $ y = \sqrt{ ( R_E^2 - x^2 ) ( ( 1 - ( R_P / R_E )^2 ) \sin( \theta_{eq} )^2 + ( R_P / R_E )^2 } ~ ~ ~ $ see [TiltingOblate] The m288 orbit is a circle in the equatorial plane with a radius of $ R_{m288} $. This circle projects into the X,Z plane as $ y = \sin( \theta_{eq} ) \sqrt{ R_{m288}^2 - x^2 } $ Two of the four points where these y values are equal are the points were the orbit enters or leaves the eclipse, so: $ y_e = \sin( \theta_{eq} ) \sqrt{ R_{m288}^2 - x_e^2 } = \sqrt{ ( R_E^2 - x_e^2 ) ( ( 1 - ( R_P / R_E )^2 ) \sin( \theta_{eq} )^2 + ( R_P / R_E )^2 } $ Let's solve for $ x_e $: $ \sin( \theta_{eq} )^2 ( R_{m288}^2 - x_e^2 ) = ( R_E^2 - x_e^2 ) \left( ( 1 - ( R_P / R_E )^2 ) \sin( \theta_{eq} )^2 + ( R_P / R_E )^2 \right) $ $ x_e^2 \left( \left( ( 1 - ( R_P / R_E )^2 ) \sin( \theta_{eq} )^2 + ( R_P / R_E )^2 \right) - \sin( \theta_{eq} )^2 \right) ~ = ~ R_E^2 \left( ( 1 - ( R_P / R_E )^2 ) \sin( \theta_{eq} )^2 + ( R_P / R_E )^2 \right) - \sin( \theta_{eq} )^2 R_{m288}^2 $ $ x_e^2 \left( ( R_P / R_E )^2 ( 1 -\sin( \theta_{eq} )^2 ) \right) ~ = ~ R_P^2 - ( R_{m288}^2 + R_P^2 - R_E^2 ) \sin( \theta_{eq} )^2 $ $ \large x_e ~ = ~ \sqrt{ { R_P^2 - ( R_{m288}^2 + R_P^2 - R_E^2 ) \sin( \theta_{eq} )^2 } \over { ( R_P / R_E )^2 ( 1 -\sin( \theta_{eq} )^2 ) } } $ Which simplifies to the following: || 6,378,137 m || $ R_E $ || Equatorial Radius || || 6,356,752 m || $ R_P $ || Polar Radius || || 12,788,866 m || $ R_{m288} $ || m288 Orbit radius || || 14,400.0 sec || $ T_{orbit} $ || sun-relative period || || 23.439281° || $ \phi $ || axial tilt || || 0° to 360° || $ \beta $ || time of year from spring equinox || $ \Large x_e ~ = ~ R_E \sqrt{ { 1 - ( ( R_{m288}^2 + R_P^2 )/ R_E^2 ) - 1 ) \sin( \theta_{eq} )^2 } \over { 1 -\sin( \theta_{eq} )^2 } } ~ \approx 6,378,137 m \sqrt{ { 1 - 0.635085 \sin( \beta )^2 } \over { 1 - 0.158227 \sin( \beta )^2 } } ~ ~ ~ \sin( \theta_{eq} ) = \sin( \beta ) \sin( \phi ) $ The eclipse fraction $ F_E $ (fraction of the total orbit) is: $ \large F_E = \arcsin( x_e / R_{m288} ) / \pi ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ $ assumes arcsin() in radians, if in degrees divide by 180° The eclipse time $ T_E $ is $ \large T_E = T_{orbit} \arcsin( x_e / R_{m288} ) / \pi ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ $ $ x_e $ varies from 4,199,446 to 6,378,137 meters, so the fraction-of-orbit $ F_E $ varies from 0.1065 to 0.1662, with an average of 0.1388, and eclipse time varies from 1534 to 2393 seconds with an average of 1999 seconds. The eclipse will be longest in spring and fall, shortest in summer and winter. The shortest eclipse periods correspond to surface peak power demands (heating and cooling), which is fortuitous. The cost of computing will be highest during summer in the temperate regions. Here's a plot versus time. Remember, the abscissa is degrees of the year, one degree=1.015 days . 30 degrees equals a month. The plot starts on the spring equinox, around March 23. {{ attachment:ecl01.png }} [[ ecl01.c | source ]] . . . [[ ecl01.txt | output data ]] . . . [[ ecl01.gp | gnuplot control file ]] |
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 ( 
asin(6371
12789)180
In fact, the Earth has an equatorial radius of 6378.1 km, a polar radius of 6356.8 km, and an axial tilt of 
=
The variable 
Oblate Earth
The equatorial plane is tilted towards the sun by angle 
eq
eq)=sin()sin()
The earth can be approximated as an elliptical disk, a projection of a ellipsoidal spheroid with an equatorial radius
(RE2−x2)((1−(RPRE)2)sin(eq)2+(RPRE)2
The m288 orbit is a circle in the equatorial plane with a radius of
eq)R2m288−x2
Two of the four points where these y values are equal are the points were the orbit enters or leaves the eclipse, so:
eq)R2m288−x2e=(RE2−x2e)((1−(RPRE)2)sin(eq)2+(RPRE)2
Let's solve for x_e :
\sin( \theta_{eq} )^2 ( R_{m288}^2 - x_e^2 ) = ( R_E^2 - x_e^2 ) \left( ( 1 - ( R_P / R_E )^2 ) \sin( \theta_{eq} )^2 + ( R_P / R_E )^2 \right)
x_e^2 \left( \left( ( 1 - ( R_P / R_E )^2 ) \sin( \theta_{eq} )^2 + ( R_P / R_E )^2 \right) - \sin( \theta_{eq} )^2 \right) ~ = ~ R_E^2 \left( ( 1 - ( R_P / R_E )^2 ) \sin( \theta_{eq} )^2 + ( R_P / R_E )^2 \right) - \sin( \theta_{eq} )^2 R_{m288}^2
x_e^2 \left( ( R_P / R_E )^2 ( 1 -\sin( \theta_{eq} )^2 ) \right) ~ = ~ R_P^2 - ( R_{m288}^2 + R_P^2 - R_E^2 ) \sin( \theta_{eq} )^2
\large x_e ~ = ~ \sqrt{ { R_P^2 - ( R_{m288}^2 + R_P^2 - R_E^2 ) \sin( \theta_{eq} )^2 } \over { ( R_P / R_E )^2 ( 1 -\sin( \theta_{eq} )^2 ) } }
Which simplifies to the following:
6,378,137 m |
R_E |
Equatorial Radius |
6,356,752 m |
R_P |
Polar Radius |
12,788,866 m |
R_{m288} |
m288 Orbit radius |
14,400.0 sec |
T_{orbit} |
sun-relative period |
23.439281° |
\phi |
axial tilt |
0° to 360° |
\beta |
time of year from spring equinox |
\Large x_e ~ = ~ R_E \sqrt{ { 1 - ( ( R_{m288}^2 + R_P^2 )/ R_E^2 ) - 1 ) \sin( \theta_{eq} )^2 } \over { 1 -\sin( \theta_{eq} )^2 } } ~ \approx 6,378,137 m \sqrt{ { 1 - 0.635085 \sin( \beta )^2 } \over { 1 - 0.158227 \sin( \beta )^2 } } ~ ~ ~ \sin( \theta_{eq} ) = \sin( \beta ) \sin( \phi )
The eclipse fraction F_E (fraction of the total orbit) is:
\large F_E = \arcsin( x_e / R_{m288} ) / \pi ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ assumes arcsin() in radians, if in degrees divide by 180°
The eclipse time T_E is
\large T_E = T_{orbit} \arcsin( x_e / R_{m288} ) / \pi ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
x_e varies from 4,199,446 to 6,378,137 meters, so the fraction-of-orbit F_E varies from 0.1065 to 0.1662, with an average of 0.1388, and eclipse time varies from 1534 to 2393 seconds with an average of 1999 seconds. The eclipse will be longest in spring and fall, shortest in summer and winter. The shortest eclipse periods correspond to surface peak power demands (heating and cooling), which is fortuitous. The cost of computing will be highest during summer in the temperate regions.
Here's a plot versus time. Remember, the abscissa is degrees of the year, one degree=1.015 days . 30 degrees equals a month. The plot starts on the spring equinox, around March 23.
source . . . output data . . . gnuplot control file