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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

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