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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Dip and the refractive invariant**

**From:**Paul Hirose

**Date:**2013 Apr 19, 14:50 -0700

One way to calculate dip is to use the "refractive invariant", a principle of optics which says n * r * cos dip is constant at all points on a light ray. (Usually the expression uses sine of z, the zenith distance. I rewrote that to cosine of dip.) n = refractive index of the air = ca. 1.00028 at sea level r = distance from the center of Earth At the point where a ray is tangent to the sea, dip is zero. Since the cosine of 0 is 1, the invariant there is n * r. What is the refractive index n? At standard atmosphere conditions of 15° C and 1013.25 millibars, it's 1.000277523. And at sea level r is simply Earth's radius, which I will call 6371000 meters. The refractive invariant n * r * cos dip at the tangent point (the sea horizon) is therefore 1.000277523 * 6371000. That's also the value of n * r * cos a at the observer. Assume height of eye is 10 meters, and a standard atmosphere. Then refractive index at 10 meters is 1.000277258. Equate the invariant at the horizon (left side) to the value at the observer: 1.000277523 * 6371000 = 1.000277258 * (6371000 + 10) * cos dip rearranging the equation gives (1.000277523 / 1.000277258) * (6371000 / 6371010) = cos dip and solving on a calculator gives 5.55' = dip. This agrees with the almanac table for 10 m height of eye. Because both quotients are very close to 1, a more elegant and interesting solution is possible. Let's look at (6371000 / 6371010) first. After rewriting it in terms of Earth radius Re and height above sea level h, the expression is Re / (Re + h). Divide numerator and denominator by Re to get 1 / (1 + h/Re). It appears that does nothing but make the expression more complicated. But note that h/Re, the height of eye in Earth radii, is a tiny number. When you take the reciprocal of (1 + tiny number), the result is almost exactly (1 - tiny number). For example, try 1 / 1.001. Note that it's almost exactly .999. The closer to 1, the more accurate this approximation. What this means is that Re / (Re + h) simplifies to 1 - h, where h is height of eye in units of Earth radii. A similar technique can simplify the term (1.000277523 / 1.000277258). The numerator (index of refraction at the sea surface) we will call n0, and let dn be the increase in refractive index at the observer. In this example it's 1.000277258 - 1.000277523 = -2.65e-7 (that is, -2.65 times 10 to the -7). Then the expression may be written as n0 / (n0 + dn). As before, when you divide both sides of the fraction by the numerator, the result is the reciprocal of (1 + tiny number), so the quotient of the refractive indices is practically 1 - dn. (Because the index of refraction at the observer is less than at the sea surface, dn is negative.) By simplifying both quotients in this manner, the equation becomes (1 - dn) * (1 - h) = cos dip Multiplying the terms on the left yields 1 - dn - h + (dn * h) = cos dip where h = 10 m / 6371000 m = 1.57e-6 dn = 1.000277258 - 1.000277523 = -2.65e-7 The term (dn * h) is very small (about 4e-13) compared to the other two terms, so ignore it. Then the formula becomes 1 - (dn + h) = cos dip Or in this example, 1 - 1.31e-6 = cos dip Apply one more approximation. If the cosine of an angle is 1 - (tiny number), then the angle itself (in radians) is very close to √(2 * tiny number). So a good approximation of dip is √(2 * 1.31e-6) = .00162 radians = 5.56', which agrees with the almanac table. Combining all the above, dip (radians) = √(2 * (dn + h)) where dn is the difference in the refractive indices of the air at the sea surface and at the observer (negative if the latter is less), and h is the height of eye in earth radii. In minutes of arc, dip = 3438′ * √(2 * (dn + h)). Test that equation at 40 m height of eye. Assume standard atmosphere. Sea level refractive index = 1.000277523 as before. At 40 m it's 1.000276462. The difference = dn = -1.06e-6. Height of eye = h = 40 / 6371000 = 6.28e-6. Dip = 3438′ * √(2 * (-1.06e-6 + 6.28e-6)) = 11.1′. This agrees perfectly with the almanac table. In a vacuum, the refractive index is exactly 1 at any point, so dn equals zero. The formula simplifies to 3438′ * √(2 * h). That's the formula for geometric dip. In other words, the refractive invariant is true even when there is no refraction! REFERENCES I obtained refractive index values from this online calculator at NIST, which implements the Ciddor formulas: http://emtoolbox.nist.gov/Wavelength/Ciddor.asp For my computations I changed the wavelength of light to 550 nanometers (the approximate peak sensitivity of the eye) since the default is the red He-Ne laser wavelength. The calculator requires air pressure and temperature, which I obtained from this Javascript standard atmosphere calculator: http://www.luizmonteiro.com/StdAtm.aspx Andrew T. Young discusses the refractive invariant: http://mintaka.sdsu.edu/GF/explain/atmos_refr/invariant.html -- I filter out messages with attachments or HTML.