A friend mentioned several weeks ago that he'd wondered for years what the equation would be for the additional night sky afforded by elevation in height (we're both amateur astronomers), and last Sunday I took a little mental vacation and began mulling over the problem, and after nine sheets of typing-paper and seven geometrical constructions I think I finally got it.
The angle rho from eye-level down to horizon, using r in the place of rho, is
r = arccos ( ( R / ( R + h ) ) ),
R = radius of, and h eye-level height above, a sphere.
That is, the formula embodies the simplest, spherical approximation to the problem—anyone wanting corrections for deviations from sphericality, or precise great-circle-route calculations, or corrections for local gravitational (and therefore space-time-curvatural, and therefore metric-mensural) deviations, can do it themselves, and God bless 'em.
The reverse equation of course is
h = R * ( ( 1 / cos ( r ) ) - 1 ).
And the first partial derivatives are, using d in the place of delta,
dr/dh = R / ( ( h^4 + 4Rh^3 + 3R^2h^2 + 2R^3h )^(1/2) )
dh/dr = ( R sin ( r ) ) / cos^2 ( r ) = ( R tan ( r ) ) / cos ( r ) = R tan ( r ) sec ( r ).
The key to the derivation of the first equation is the geometric construction showing that the angle from an observer's eye-level down to his horizon is equal to that between his and his horizon's radii or perpendiculars: (1) Draw a circle in the middle ninth of the geometric construction space and put a heavy dot at its center; this represents of course the sphere on which the observer stands. (2) Draw a line vertically up from that dot and through the circle to the edge of the space and end it with an arrow-tip; this represents the observer's horizon and its radius and its extension. (3) Draw another line at 45 degrees to the right (at the clockwise 1:30 position) from the center dot through the circle and to the edge of the space and end it with an arrow-tip; this represents the observer's radius and its extension. (4) Draw the tangent to the horizon, a horizontal line intersecting the intersection of the horizon radius and line with the circle, extending the tangent to both sides of the space and ending both ends with arrow-tips. (5) Draw the tangent to the observer, a line intersecting the intersection of the observer's radius and line with the circle, extending the tangent to both sides of the space and ending both ends with arrow-tips. (6) Draw the observer's eye-level line intersecting the intersection of the observer's radius and line with the horizon tangent, extending the line to both sides of the space and ending both ends with arrow-tips. (7) Draw small squares around the intersections of the horizon tangent with the circle and the observer's eye-level line with the other two lines, with the top and bottom or inner and outer sides of the squares parallel to the tangent and the line respectively. (8) Darken the circle and the dot, the horizon and observer radii from the dot to the circle, the horizon tangent from the intersection with the circle to its intersection with the observer's line, the observer eye-level line, and the squares; these are the critical elements of the construction. (9) Mark the horizon and observer radii each "R", and the observer radius extension to the intersection with the eye-level line "h". (10) Mark the angle between the horizon and observer radii "(rho)"; mark the angle between the horizon tangent and the observer line "90 - (rho)"; and mark the angle between the observer eye-level line and the horizon tangent "(rho)".
Note, then, that the angle between the horizon and observer radii is equal in magnitude to that between the observer eye-level and his horizon.
Now all we have to do is apply elementary trigonometry to the former angle and we have our equation: In terms of the classic trigonometric variables, then, the horizon radius is "x" ( = R ), the observer radius is "r" ( = R + h ), and the cosine of the angle between the two, "(rho)", is of course given by "x/r" ( = ( R / ( R + h ) ) ):
cos ( (rho) ) = R / ( R + h ).
(rho) = arccos ( R / ( R + h ) ).
Running the equations by hand, taking the radius of the Earth to be 4000 miles, and using the trig tables in my beloved Riddle's Calculus and Analytic Geometry (3d ed.), for a few values, first forward, and then backward, gave (approximate) results combined in the following table:
h (mi.) rho (deg.)
Note that at 19530 miles, Earth will be 2 * ( 90 - 80 ) = 20 degrees wide.
I think the above basically met the question, and was certainly fun in the runnin'.
Keywords: armchair astronomy