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I agree with Marcel's calculation of April 23 that the traditional formula

dip = 1.76′ * √h  (eq. 1)

implies dn = -2.59e-8 * h, where dn = refractive index at eye -
refractive index at horizon, and h = height of eye (meters).

http://www.fer3.com/arc/m2.aspx/Dip-Temperature-Gradient-Tschudin-apr-2013-g23661

For an explanation of dn, see my messages on the refractive invariant:
http://fer3.com/arc/m2.aspx/Dip-refractive-invariant-Hirose-apr-2013-g23613
http://fer3.com/arc/m2.aspx/Dip-refractive-invariant-Hirose-apr-2013-g23652

I also agree that a temperature gradient of .026 °C per meter will
produce that dn. However, temperature must *increase* with height
because dn is negative. That is quite different from the -.0065 °/m of
the standard atmosphere. Also, it does not include the effect of air
pressure.

At the small heights of dip computations, pressure has practically
linear variation with height and is independent of the temperature lapse
rate. If P0 = sea level pressure, T0 = sea level temperature (C), and h
= height in meters,

P = P0 * (1 - h * .034163 / (273.15 + T0))  (eq. 2)

At h = 100 meters the approximation is accurate to 1 part in 10000.

[Originally I showed the steps by which it is derived from the standard
atmosphere formula, but deleted that part for brevity. In
http://en.wikipedia.org/wiki/Barometric_formula, equation 1, divide both
sides of the fraction in brackets by the numerator to get a reciprocal.
Invert the reciprocal and negate the exponent. Finally, transform
exponentiation into multiplication with the approximation (1+x)^y =
1+xy, which is accurate when x is near 0. Note that temperature lapse
rate Lb cancels itself after the last transformation.]

If T0 = 10 C, the formula is

P = P0 * (1 - 1.2065e-4 * h)  (eq. 3)

For small variations in air density, refractivity is almost exactly
proportional to density, so if n0 = refractive index at sea level, dP =
increase in air pressure, dT = increase in air temperature,

dn = (n0 - 1) * (dP/P0 - dT/T0)  (eq. 4)

At T0 = 10 C and P0 = 1010 mb, n0 = 1.000281622 at sea level, according
to the NIST calculator (550 nm wavelength). As I said at the beginning
of this message, equation 1 implies dn = -2.59e-8 per meter. The value
of dP/P0 may be obtained from my approximate formula for air pressure:
-1.2065e-4 per meter. When we solve for dT (the only unknown in equation
4) we get the temperature lapse rate implied by the traditional dip
formula: -.00811 °C per meter.

The quantity dn is the height of eye correction (in units of Earth
radii) due to refraction. When dn in converted to meters, I call it dn′:

dn′ = 6371000 m * .000281622 * (-1.2065e-4 * h - .00353 * dT)

dn′ = -.2165 * h - 6.337 * dT  (eq. 5)

For example, at 30 m height of eye, assuming the temperature lapse rate
I calculated, dT = -.00811 * 30 = -.24 C, and dn′ = -4.95 m.

In a previous message I showed that

dip = 1.926′ * √(dn′ + h)  (eq. 6)

so in this example at 30 m,

dip = 1.926′ * √(-4.95 + 30) = 9.64′

Equation 1, the traditional formula, agrees to better than .01′.

If air temperature at observer is identical to sea level, equation 5
says dn′ = -6.50 m, and equation 6 says dip = 9.34′. So a quarter degree
C changed dip .3′. Temperature gradients may be a significant source of
error in dip measurements from land, or even at sea if the vessel is a
"heat island".

Note that equations 3, 5, and 6 assume the standard nautical almanac
atmosphere. Even equation 6 can change a little, depending on the
assumed mean radius of Earth.

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