#format jsmath
= Tilting the Oblate Earth =

The earth is an oblate spheroid, with an equatorial radius $ R_E $ = 6,378,137 meters and a polar radius  $R_P$ = 6,356,752 meters ( [[ http://en.wikipedia.org/wiki/World_Geodetic_System | GPS coordinates ]] ).  It projects an ellipse when viewed from the equatorial plane, and a circle when viewed from the poles.  In between, it also presents an ellipse.  What is the shape of that ellipse?

Our reference directions are x to the right, y to the back, and z to the top.  When we tilt the oblate spheroid backwards, the front moves up and the back moves down.   We can slice the spheroid into a series of elliptical slices along the x axis, each slice in a plane in y and z, perpendicular to the x axis.  By analyzing the vertical projection of each slice, we can generalize to the vertical projection of the whole spheroid. 

The equation for the ellipsoid is

$ 1.0 = x^2/RE^2 + y^2/R_E^2 + z^2/R_P^2 $

The equation for the elliptical slice through the center of the spheroid is:

$ 1.0 = y^2/R_E^2 + z^2/R_P^2 ~ ~ ~ x = 0 $  or alternately,  $ x = R_E \cos( \theta ) $ and $ y = R_P \sin( \theta ) $.  We can rotate that ellipse by $ \theta_{eq} $:

$ [ x_{rot}, y_{rot} ] = \left( \begin{array}{cc} \cos( \theta_{eq} ) & -\sin( \theta_{eq} ) \\  \sin( \theta_{eq} ) & \cos( \theta_{eq} ) \end{array} \right) \times [ R_E \cos( \theta ), R_P \sin( \theta ) ] $

$ [ x_{rot}, y_{rot} ] = [ R_E \cos( \theta_{eq} ) \cos( \theta ) - R_P \sin( \theta_{eq} ) \sin( \theta ) ~ , ~ R_E \sin( \theta_{eq} ) \sin( \theta ) + R_P \cos( \theta_{eq} ) \cos( \theta ) ] $
 
All we need to know is the maximum of $ y_{rot} = R_E \sin( \theta_{eq} ) \sin( \theta ) + R_P \cos( \theta_{eq} ) \cos( \theta ) $ as a function of $ \theta $. 
Differentiating the above, we get 

$ d y_{rot} / d \theta  = R_E \sin( \theta_{eq} ) \cos( \theta ) - R_P \cos( \theta_{eq} ) \sin( \theta ) $.  

Setting that to zero, we get:  

$ R_E \sin( \theta_{eq} ) \cos( \theta ) = R_P \cos( \theta_{eq} ) \sin( \theta ) $

and so:

$ \sin( \theta ) / \cos( \theta ) = tan( \theta )  = ( R_E / R_P ) ( \sin( \theta_{eq} ) / \cos( \theta_{eq} ) = ( R_E / R_P ) \tan( \theta_{eq} ) $

$ \sin( \theta ) = \tan( \theta ) / \sqrt{ \tan( \theta )^2 + 1 } $

$ \cos( \theta ) =              1 / \sqrt{ \tan( \theta )^2 + 1 } $

$ y_{rot, max} = \left( R_E \sin( \theta_{eq} ) \tan( \theta ) + R_P \cos( \theta_{eq} ) \right) / \sqrt{ \tan( \theta )^2 + 1 } $

$ y_{rot, max} = \left( R_E \sin( \theta_{eq} ) R_E \sin( \theta_{eq} ) / R_P \cos( \theta_{eq} ) + R_P \cos( \theta_{eq} ) \right) / \sqrt{ \tan( \theta )^2 + 1 } $

$ y_{rot, max} = ( R_E^2 \sin( \theta_{eq} )^2 + R_P^2 \cos( \theta_{eq} )^2 ) / ( R_P \cos( \theta_{eq} ) \sqrt{ R_E^2 \sin( \theta_{eq} )^2 / R_P^2 \cos( \theta_{eq} )^2  +  1 } ) $

$ y_{rot, max} = ( R_E^2 \sin( \theta_{eq} )^2 + R_P^2 \cos( \theta_{eq} )^2 ) / \sqrt{ R_E^2 \sin( \theta_{eq} )^2 + R_P^2 \cos( \theta_{eq} )^2 } $

$ y_{rot, max} = \sqrt{  R_E^2 \sin( \theta_{eq} )^2 + R_P^2 \cos( \theta_{eq} )^2 } $

$ y_{rot, max} = \sqrt{ ( R_E^2 - R_P^2 ) \sin( \theta_{eq} )^2 + R_P^2 } $

and finally, as a function of x, the projected ellipse for the oblate earth is:

$ y = sqrt{ 1 - x^2 / R_E^2 } \sqrt{ ( R_E^2 - R_P^2 ) \sin( \theta_{eq} )^2 + R_P^2 } $

$ \Large y = sqrt{  ( R_E^2 - x^2 ) ( (  1 - ( R_P / R_E )^2 ) \sin( \theta_{eq} )^2 + (  R_P / R_E )^2 } $

Anything behind this ellipse will be eclipsed.