NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Dip and Temperature Gradient
From: Richard B. Langley
Date: 2013 May 14, 17:44 -0300
From: Richard B. Langley
Date: 2013 May 14, 17:44 -0300
For those interested in how surveyors may have handled refraction (including temperature gradients) in high precision surveys, here are a couple of links to a UNB thesis and a contract report on the subject. Note that the thesis is not from one of my students but that of a department colleague, now Prof. Em. Adam Chrzanowski: http://gge.unb.ca/Pubs/TR132.pdf http://gge.unb.ca/Pubs/TR142.pdf Prof. Chrzanowski is still around and active should anyone want to pose a question to him. -- Richard Langley Quoting "Bruce J. Pennino": > 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 Hirose > To: bpennino.ce---net > Sent: Monday, May 13, 2013 5:54 PM > Subject: [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 > > > : http://fer3.com/arc/m2.aspx?i=124039 ----------------------------------------------------------------------------- | Richard B. Langley E-mail: lang@unb.ca | | Geodetic Research Laboratory Web: http://www.unb.ca/GGE/ | | Dept. of Geodesy and Geomatics Engineering Phone: +1 506 453-5142 | | University of New Brunswick Fax: +1 506 453-4943 | | Fredericton, N.B., Canada E3B 5A3 | | Fredericton? Where's that? See: http://www.fredericton.ca/ | -----------------------------------------------------------------------------