NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Dip and Temperature Gradient
From: Bruce J. Pennino
Date: 2013 May 14, 12:13 -0400
From: Bruce J. Pennino
Date: 2013 May 14, 12:13 -0400
Bruce
Thank you Paul for helping with this.
I've recorded more dip data from land locations and
I'll eventually let everyone see it. I've been told that surveyors were
supposed to take critical precise measurements in the middle of the day when
refraction was less significant. Maybe temperature and pressure gradients
over land were less severe. Agree? Furthermore the issue of temperature and
pressure gradients cannot be thought about without considering wind. The
turbulence and mixing created by the wind alters the equations. When you
consider the boundary layer, particularly at low elevations mixing becomes an
issue. Even though a boat or a beach can be a local heat source/sink, the
observer's eyes are high enough "to feel the wind in your face". I feel the
theodolite shaking. We all know it is even windier on the ocean. For
modest HoE, the air and pressure gradient mixing at the observer and the
horizon is more or less the same excepting on very hot days. I would not
set up in a parking lot overlooking a cold ocean, or a blazing hot sandy
beach . Getting representative data is tough.
I'm going to repeat some of my previous dip
measurements on cloudy, temperate days (if possible). I'm hoping to set up
on damp sand just after high tide for one set of data. Then move higher up
on dune/ beach to get another set of data. I will purchase a
thermometer to get a " local reading somewhere nearby" in
shade .
I also know an elevated place, El 100 +/- MSL
, where I can set up in the shade of an overhanging roof and then quickly
move to a somewhat lower elevation in the sun. I know these
elevations to within +/- 1.5 ft related to MSL.. Open to thoughts on places
to set up. I normally record in my field notes the wind and cloud
conditions.
Regards
Bruce
----- Original Message -----From: Paul HiroseSent: Monday, May 13, 2013 5:54 PMSubject: [NavList] Re: Dip and Temperature Gradient
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. --: http://fer3.com/arc/m2.aspx?i=124035
