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    Re: Measuring (and calculating) Dip
    From: Frank Reed
    Date: 2013 Feb 28, 16:35 -0800

    We've been over this many times in NavList messages, though I suppose it's been a few years. Time flies!! :)

    There are variety of equations that we can generate with some basic geometry for light rays under the assumption that they are perfect straight lines, and the Earth is a perfect sphere. We can calculate dip, distance to the horizon, distance by vertical angle, distance of an object peeking above the horizon, and a few other things. There are tables and cases in the back of Bowditch. But let's stick with dip.

    If you draw yourself a triangle, you can easily see that the dip of the horizon is given by


    where h is the observer's height and R is the actual radius of the Earth. First of all, let's drop the cosine. The cases that will interest us below will always involve small angles, and unless we're in high orbit around the Earth, the difference never matters. This equation, for small angles, reduces to


    Just to be clear here, it is NOT better to keep the "arccos". Doing so creates a misleading impression of what we can actually calculate. To derive this equation from the cosine version, remember that cos(x)=1-x^2/2 and also remember that 1/(1+x)=1-x for relatively small values of x (in both).

    Now, light rays near the surface of the Earth do not obey the laws of geometry. They obey the laws of physics. But in this case, we get to use fairly easy physics and limit ourselves to simple refraction. This is a refraction problem that most people never encounter (but it's not rocket science --just plain old refraction). It's not a lens or prism where a light ray enters a denser medium at some angle. The air of the atmosphere is stratified by density. Air closer to the ground is denser than air higher up. In all of the cases of interest to us in terms of dip, dip short, etc., we are dealing with light rays that are travelling nearly horizontally. What happens to a ray of light travelling horizontally in a medium that is denser at lower levels? Simple ray optics might lead you down the wrong path. A light ray travelling perfectly level would stay at one level of air density. Why should it deflect; it's not crossing any density boundaries? But if you think in terms of wavefronts of light and remember that higher density air implies slightly lower speed of light, you can see that the wavefronts will be turned because their lower "edges" are in denser air and therefore they move more slowly. So the direction of the light's travel, which is always perpendicular to the wavefronts, will bend downward in normal, density-stratified air. And the greater the density gradient, the more downward bending of the light's trajectory. We can quantify that bending.

    First let's deal with units. Suppose a light ray is travelling directly towards the horizon. Suppose the air bends its trajectory downward by a certain number of minutes of arc in one nautical mile. Perhaps it is deflected downward by 0.1' in a nautical mile. What are the units of that deflection? Well, if we remember that a nautical mile is a minute of arc as measured from the center of the Earth, then it turns out that this rate of deflection is dimension-less. A deflection rate of 0.1' per nautical mile is equal to 0.1 minutes of arc per minute of arc or simply 0.1. There are some differences in conventions, but some multiple of this is the dimension-less constant k or beta0 that you will find in references on this topic. I'll call it k.

    If the dimension-less quantity k is 0, this means that a light ray is not deflected at all by terrestrial refraction. This would be the case (unstable in Nature) where the atmosphere has a constant density. If k=0, light rays are not bent at all, and we can use the pure geometric expressions for dip and other similar quantities. This is rarely the case. It's also possible for the atmosphere to have such a high density gradient, that light rays bend downward at a rate of 1' per nautical mile, implying k=1.0 or even higher. This is also rare, but the implications of that exact value are fun to consider. If a light ray fired horizontally from some point on the Earth bends downward at that rate, k=1, then it will orbit the Earth, following the Earth's curvature at a rate of 1 minute of arc deflection for every nautical mile of horizontal travel. Under such conditions, light rays coming from any distant objects will behave as if the Earth is perfectly flat. You can see forever! In reality, extinction of light limits this "forever vision" to some miles before it fades into a visual "fog", but optical measurements of any objects visible before extinction wins out will all be consistent with the objects resting on a flat plane. The Earth will look flat.

    This downward deflection of light rays is geometrically equivalent to a flat geometry problem where we simply reduce the curvature of the Earth's surface. If all light rays travelling more or less horizontally are deflected by k minutes of arc per nautical mile, then it is just as if the Earth has been flattened by that quantity k. Instead of bending every light ray downward, we can just bend the surface of the Earth upward (mathematically, that is). This can be proved mathematically if you're so inclined, but you can see the general concept just by drawing a little picture. Flattening the Earth by some fraction is, of course, equivalent to increasing its radius by a similar amount. So we can take any equation that we have derived by pure geometry and FULLY INCORPORATE this basic kind of refraction by replacing the radius of the Earth R with some equivalent radius Re with Re=R/(1-k). This is how you can derive the actual value of dip and the various other relationships of terrestrial refraction (such as those tabulated in the back of Bowditch).

    But then there's weather. When you spend your days contemplating the perfectly predictable motions of the celestial bodies, it can be a bit annoying to realize that there are things that are unpredictable in celestial navigation observations, and not just observational noise, but actual physical phenomena that we cannot predict. That quantity k, above, depende on the density gradient off the atmosphere. The pressure gradient of the atmosphere is tightly constrained by hydrostatic equilibrium: the air pressure at sea level has to support the weight of the entire column of air above it. But alas, density also depends on temperature. If the air gets colder with altitude at a rapid rate, the density of the air can be constant with altitude. But normally, the air gets colder with altitude at a rate near the "moist adiabatic" rate. It doesn't have to. That's just the norm. The air can, and frequently does, get warmer with altitude. This is called a temperature inversion. In a temperature inversion, the air gets less dense with altitude at a much higher rate. That higher density gradient increase the value of k, sometimes allowing it reach 1 or even higher. If the rate of change of temperature is -34.1°C/km then k=0. If this "lapse rate" is about +110°C/km, then k=1. The average atmospheric state is between these numbers --a rate of -7°C/km is about normal, and usually k=0.16 or so.

    We can also express k in terms of the atmosphere's "scale height". The scale height is the altitude at which the density decrease by a factor of "e". You can also call it the "e-folding" height. It's typically around 11 kilometers. The scale height, s, is related to the lapse rate by s=R*T/(g+Lr*R) where R is a constant, g is the acceleration of gravity, T is the (absolute) temperature, and Lr is the lapse rate. The key feature here is that the scale height is more or less inversely proportional to the lapse rate. The deflection rate k is approximately given by k=Q*(1.8km)/s where Q is the usual factor for non-standard sea level pressure and temperature.

    Many theoreticians have re-discovered this physics in the past 150 years or so and reached the same erroneous conclusion. They imagine that we can calculate k by measuring actual temperature differences. And thus we could eliminate much of the uncertainty in dip and related quantities. It hasn't worked out that way. There's a lot of variability in the lower atmosphere that can modify the dip in ways that are not accounted for by a "pure" model of atmospheric lapse rate. But the lapse rate model does at least allow us to place some limits on possible ranges. We can understand the origin of the variability of dip even if it does not give us a means of calculating it. Weather is just too messy for that.

    I do want to emphasize though that the analysis above which lets us replace the actual radius of the Earth with an equivalent radius does not have to be re-derived for every specific case of terrestrial refraction. That methodology applies to each and every case where we're dealing with light rays travelling close to the Earth's surface. For example, there's an equation underlying "Table 15" in modern editions of Bowditch that has been a topic of discussion before. It's a means of determining "distance by vertical angle measured between sea horizon and top of object beyond sea horizon". The equation in the explanation in Bowditch looks a little exotic, but you can derive it by simple geometry for a flat Earth. Then anywhere you find R, replace it by Re. You will discover when you do this, that the various tables in Bowditch have been created with somewhat different values for k. This is inconsistent but not really "wrong". The bigger problem is that the exact values for the constants in the equations imply accuracy that is not real.


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