Navigators use the term "dip" to refer to the effect of height of eye above sea 
level has on the altitude of a celestial body above the apparent horizon. 
Similarly , the distance to horizon is greater at higher eye elevations. A 
sailor on ship deck has an advantage in spotting land over one at sea level; 
but the sailor in the crow's nest
sees land first.

Any nautical celestial navigation book will have a table. The Nautical Almanac 
also.
The simple formula is D = 1.169 SQR(h) where D is distance to horizon in 
nautical miles and h = height of observers eye in feet.  [ If using meters the 
constant is 2.07]

Use of this simplified formula is usually all that is needed. It requires of 
course a clear day to the horizon, perhaps questionable in Great Britain     :)

For greater accuracy try:
            D= SQR( [2*r*h] / [6076.1 *B] )

That  'B' will be a beta in the books. Again:     D is distance to the horizon 
in nautical miles; r is the mean radius of the earth ( 3440.1 nautical miles) ; 
 h is the height of eye IN FEET; and B is a contant relating to terrestrial 
refraction, 0.8279

Seems to me that B could vary depending on atmospheric conditions but I never 
studied it to that detail. However, Bowditch, "The American Practical 
Navigator" (a classic tome) states the " error in refraction is generally less 
than that introduced by nonstandard atmospheric conditions"

Good Luck

DAVE

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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