The prime vertical radius of curvature of the ellipsoid (the radius of the
osculating circle in the east-west direction) is given by
a^2
N = -------------------------------------------
(a^2 * (cos phi)^2 + b^2 * (sin phi)^2)^0.5
The ellipsoid of the Geodetic Reference System 1980 has
a = 6 378 137 metres
b = 6 356 752 metres (to the nearest metre).
In computing the distance to the horizon, if you want the visibile horizon (as
opposed to the geometric one), you should take refraction into account.
-- Richard Langley
Professor of Geodesy and Precision Navigation
where a and b are the semimajor and semiminor axes of the ellipsoid and phi is
the geodetic latitude.
On Mon, 3 Aug 1998, Tony Moss wrote:
>Fellow Shadow Watchers,
> Information boards are being prepared for a
> nearly-completed large dial on a hill near the Northumberland (UK) coast
> which is 94 metres above mean sea level at Latitude 55 1' 38" North
> Longitude 1 30' 16" West.
>
> The latest question is "How far away is the sea horizon?" School geography
> taught me that the earth is an 'oblate speroid' so I suppose the true
> distance varies slightly depending on the direction in which the observer is
> looking but so little as to be unimportant perhaps? The sea is only visible
> in a generally easterly direction.
>
>Can any list member supply the mean sea level radius of the earth at this
>location on which to base the necessary trig calculation plus any subtleties I
>may have overlooked as I don't have ready access to specialist reference
>material of this sort.
>
>With thanks in anticipation of any helpful response.
>
>Tony Moss
>
>
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Richard B. Langley Internet: [EMAIL PROTECTED] or
[EMAIL PROTECTED]
Geodetic Research Laboratory BITnet: [EMAIL PROTECTED] or
[EMAIL PROTECTED]
Dept. of Geodesy and Geomatics Engineering Phone: (506) 453-5142
University of New Brunswick FAX: (506) 453-4943
Fredericton, N.B., Canada E3B 5A3 Telex: 014-46202
Fredericton? Where's that? See: http://www.city.fredericton.nb.ca/
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