On approaching a coast from the sea, it frequently happens that the first sight of land is the upper part or top of a mountain whose outline is dimly but distinctly seen against the sky. Possibly this particular mountain is the landfall expected, and the height already noted. At any rate, the dead-reckoning position will generally be sufficiently accurate to make the identification, and it only remains to take the sextant altitude above the horizon and compute the distance. Under normal atmospheric conditions this, together with the compass direction of the summit, gives an accurate fix.
The formula for determining the distance of a mountain peak that lies beyond the horizon is simple; but its solution is too tedious to make it of practical navigational value.
In the March, 1933, issue of the U. S. Naval Institute Proceedings, an excellent article by Lieutenant Commander A. F. France, U. S. Navy, gives a formula published some years ago by Captain G. L. Schuyler, U. S. Navy, and also an excellent method for constructing a graph derived from the formula. The co-ordinates of the graph are sextant altitude of peak and distance in nautical miles. A series of curves are drawn representing various heights of mountains from 500 feet to 6,000 feet, at intervals of 500 feet. These are computed for a height of eye of 27 feet.
Such a graph is particularly valuable for smaller vessels operating offshore in tropical or semi-tropical regions where barrier reefs make it essential to have an accurate fix when approaching land. The accompanying graph is therefore computed for height of eye of 9 feet above the sea, and may be used with equal facility by either small or large vessels. The derivation of the formula may be of interest.
Let D = distance in nautical miles from observer to top of mountain
h = height of eye above sea, in feet
m = observed angle in minutes of arc between horizon and distant peak
H = height of mountain, in feet
R = mean radius of the earth (taken at 20,890,590 ft.)
c = coefficient of terrestrial refraction (taken at 0.07269)
Sine of one minute of arc =0.00029089
Nautical mile = 6,080 ft.
Also, it is assumed that the terrestrial curvature of light is the arc of a circle whose radius is constant within the limits of this problem. The errors induced from this assumption are not sufficiently great to be considered.
Some years ago the U. S. Coast and Geodetic Survey by observations in New England found the coefficient of refraction to vary in value between 0.071 and 0.078. The mean of 137 observations on the New York State survey gave a value of 0.073. In view of the considerable variation of refraction at sea level, it is safe to assume a value, c = 0.073, as being a close approximation to an average. However, in the following formula a value for c is used equal to 0.072686—a figure which simplifies the final equation without varying the results by a measurable amount, the difference being less than one in one thousand.
To Determine Distance to Horizon
AC is the arc of a circle (assumed to be the earth’s surface)
R is the radius AE
AB is the tangent at point A
AD is a perpendicular to line EB, where B is any point on tangent AB
a is angle at E between AE and EB
If we consider AC to be the surface of the sea, and if there were no refraction, then BC would be the height of eye necessary to see the point A on the horizon at a distance AB.
But light has a curved trajectory which within the limits of this problem may be considered the arc of a circle whose radius we will call Rr’
If we draw a figure similar to Fig. I in which AEr or Rr is the radius of the arc of a light ray, making AB a tangent to the ray and of the same length as AB in Fig. I then BCr is the distance that the light ray is refracted in the distance AB and we get, as in equation (3) Combining the essential features of Fig. I and Fig. II we have Fig. III. From this it is seen that Cr would be the point above the sea at (C) from which the point A could be seen at a distance of AB, i.e., CrC = height of eye (h), but CrC = BC—BCr’
Also it is evident that as we consider with very small error that the curvature of vision BT has the same radius as the curvature of vision BB'
This is the general equation from which graphs may be easily constructed for any height of eye and for any height of mountain.
While the accompanying graph is computed for height of eye of 9 feet, it may be used with sufficient accuracy for any height of eye between 6 feet and 16 feet without entailing graph reading errors greater than 1 mile.
When height of eye is 9 feet, under normal atmospheric conditions the graph is correct to within 0.2 miles.
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