# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Refraction near the Horizon ? Ob servation vs. Calculation**

**From:**Brad Morris

**Date:**2013 Apr 6, 12:05 -0400

Hi Marcel

In plotting the four curves, I wanted to see how the nominal equations

0.02977sqrt(h,meters)

0.971sqrt(h,feet)

3438sqrt(2h/R)

compared to the equation wherein pressure and temperature are inputs instead of assumed to always being nominal. I have not used dip,minutes = 1.76sqrt(h,meters). Perhaps you have confused me with Bruce?

In so far as the constants, yes, the 238K should be 283K. I transposed the digits. This has no affect on the curves, as the (P/1010)/(T/283) term was assumed to be unity or 1.

The mean radius of the earth, given by many internet sources is 6371000 meters. Many other internet resources give the radius as 6400000 meters. While this 29000 meter delta is only a difference of 1/2 of 1%, it is important to keep our constants uniform. Of course I understand the earth is not a true sphere and all of these values are an attempt to force a three dimensional ellipsoid into a true sphere.

I am somewhat puzzled by your assertion that the only affect of pressure and temperature is on the 0.3 moa and not on the geometric airless dip. In the equation R/(1-k), we change the effective radius of the earth such that geometric airless dip matches the refracted dip. When we change pressure and temperature, these only have an effect on k. Thus the minor effect of pressure and temperature is on the effective radius of the earth. The dip then calculated is as if there was no atmosphere. Did I miss something here?

I'll try it a much more simple way. Would you kindly present the equations by which you modify the dip by pressure and temperature. While it may be obvious to you, its not so obvious to me. If you would also kindly work an example by these equations, I would appreciate it.

Best Regards

Brad

P.S. 2: In relation with terrestrial refraction you find some times also an uppercase "K". This corresponds to the inverse of the lowercase "k", K=1/k and describes how many times the radius of the light ray is larger than the radius of the earth.P.S: The standard conditions for "nautical" applications are generally 10C and 1010hPa. The 10C correspond to 283.15K and not to 238K; which is possibly a typo.MarcelIt is now only this difference of 0.3 moa which would require to be adjusted to P and T. In many cases the pressure and temperature correction of the dip is so small that it can be neglected.DIPrefr = 3.5 moaDIPgeom = 3.8 moaThe geometrical dip, DIPgeom, is the angle to the horizon on an air-free earth. Because there is no refraction it is calculated with k=0 resulting ink = 0.168 (your beta)HoE = 4 mREarth = 6400000 mBrad,

It appears to me that there is a fundamental misunderstanding regarding the dip. Note that not all of the dip results from refraction. The dip consists of two parts, a geometrical angle and an angle difference resulting from refraction. If you want to apply a temperature and pressure correction to the dip you only have to do this to the difference resulting from refraction. The simplified formula DIP[moa]=1.76*SQRT(HoE) does not provide this distinction. However, calculating the dip with the terrestrial refraction value (your "beta" but mostly denominated with "k") allows to calculate the two contributions.

Example:

However, we do have an atmosphere which bends the light ray corresponding to a refraction of e.g. k=0.168:

Because of this bending of the light ray (with k=0.168) the DIP IS REDUCED (and the DISTANCE TO HORIZON INCREASED), i.e. the air shifts in this case the horizon upwards by 0.3 moa:

On Sat, Apr 6, 2013 at 1:16 AM, Brad Morris <bradley.r.morris---com> wrote:

Hi Frank

Thank you for that series of equations for standard pressure and temperature when calculating dip.

FOR NON-STANDARD PRESSURE AND TEMPERATURE:

Step 1) Record the air pressure and air temperature.

Example: 4//5/2103 3:50PM Buoy 44025 has air temp 44.1 degrees F and 29.73 inches of mercury.Step 2) Convert to millibars and degrees kelvin.

Example continued: 29.73 inches * 33.8637526 = 1013.203 millibars

(44.1-32)*5/9+273=279.722KStep 3) Solve for Q = (P/1010mb)/(T/238K)

Example cntd: Q=(1013.203/1010)/(279.722/238) = .8535430404Step 4) Solve for beta = alpha*Q*Re/s; where alpha = .000281, Re is the radius of the earth in meters 6371000, s=9000 meters

Example cntd: beta = .000281*.8535430404*6371000/9000

Beta=.1697840313Step 5) Solve for equivalent radius R= Re/(1-beta)

Example cntd: R=6371000/(1-.1679840313) = 7673906.838metersStep 6) Solve for dip = arccos(R/(R+h)) assume h = 3.5 meters

Example: dip=arccos(7673906.838/(7673906.838+3.5)) = .0547221697 degrees = 3' 16.999"NOTE: I know Frank, that you object to using the arccos here. If you wish to substitute your conversion, you may.

+++++

USING STANDARD PRESSURE AND TEMPERATURE EQUATIONS, WE HAVEdip = 0.02977*sqrt(h)

dip = .0556945702

= 3' 20.500"Or perhaps to the equation that you Frank have offered

dip(minutes)=3438*sqrt(2h/R)

=3438*sqrt(2*3.5/6371000)

=3' 36.223"++++++

Since my conditions in the example are fairly close to "standard conditions", I expected a reasonable correlation between the values.In fact, I see only 3.5" of delta dip when comparing the non-standard to standard 0.02977... That seems pretty close!

Yet I see 19.2 seconds of delta when I utilized the 3438... equation. That's 0.32 MINUTES of delta in equation alone.

++++++

So firstly, is the non-standard pressure and temperature equation for dip correct? I got that by digging in the archives and some guy named Frank Reed gave that equation. Hopefully, I got it right!Secondly, is the 0.02977sqrt(h) acceptable for the purposes of "standard pressure and temperature).

Thirdly, why does it differ from you 3438sqrt(2h/R)?

Or are the "standard p+t" equation differences really just that differing constant you have spoken of?

Brad

On Apr 5, 2013 3:12 PM, "Frank Reed" <FrankReed---com> wrote:Bruce Pennino wrote:

"I was interested to see that you use the equation for dip 1.76 sq rt H meters, which is the same as 0.971 sq rt H feet. I gather that this is the universally accepted equation, which I've proven to myself is from basic trig knowing the average radius of the earth? Right?"Not quite. The constant 0.97 has TWO sources. You can do trig, work a series expansion, and calculate. That will give you one constant (see below). But, as I always try to point out, and HAVE SAID many times in the past couple of months, that calculation is strictly a geometry result: it gives the wrong constant. Unfortunately, dip is not a geometry problem -- it's a physics problem. Light rays don't travel in straight lines. They are bent by refraction. It's a well-established physics problem, and the result under the simplest assumptions about the nature of the air turn out to be very simple: the result of the physics calculation is identical to the geometric calculation but with the actual radius of the Earth, R, replaced by an "effective" radius, R/(1-k), and this applies to any calculation involving terrestrial refraction --not just dip. I posted a diagram on this after I thought everyone here understood all of this (link below). NONE of this dip discussion will make sense to you until you understand this concept of refraction modifying the dip relationship. The average value of that quantity k, which is the downward rotation of a light ray per nautical mile is 0.16. But the reason there's variability is because k is fundamentally weather-driven. If you can precisely measure the properties of the atmosphere when you're making observations, you can calculate a more accurate value of dip. But in general, this is not possible. It's WEATHER.

The geometry for calculating dip is simple enough. I'll just quote the result for the basic triangle, which I think everyone can draw without explanation, before we narrow down from geometry to reality:

cos(dip) = R/(R+h)

where R is the radius of the Earth and h is the observer's height of eye. When you draw your triangle, you also find that the angle "dip" is ALSO equal to the angle at the center of the Earth between the observer's location and the point where a straight ray from the observer's eye just meets the Earth. That means that the dip is identical to the distance to the horizon, measured as an angle geocentrically. Or, in more practical terms, dip in minutes of arc equals distance to the horizon in nautical miles. Note carefully: this is the pure geometric solution. The equation does NOT match observations.The next step in the pure geometric solution is to limit our altitude h to something less than a mile or two. We're not interested in the horizon as seen from space at this point in time. We're also not interested in horizon calculations from five or ten miles up since the horizon is always lost to "extinction" and simple atmospheric haze from altitudes above a mile or so (and frequently much lower). In practical terms, you can't see the sea horizon from a high-altitude airplane.

Since we are limiting h to be less than a mile, the quantity h/R is a very small number. Also, if we run an exact calculation of the angle, the dip angle is a very small number as long as h is less than a mile. That means that we can use series expansions on the equation above. The series expansions we need are as follows (for the cosine, this assumes the angle is "pure ratio" or --same thing-- the angle is "in radians"):

cos(x) = 1 - x^2/2

1/(1+x) = 1 - x

These are valid approximations for any x so long as x is much less than one. We take the original equation and re-write it (now using d for dip) first by dividing the numerator and denominator on the right by R:

cos(d) = 1/(1+h/R).

We use the cosine series on the left and the binomial series on the right:

1 - d^2/2 = 1 - h/R.

Cancel the 1, reverse the signs:

d^2/2 = h/R.

And finally:

d = sqrt(2*h/R).

This version of the equation applies for any units, and its result is an angle as a pure ratio. But if we want the angle in minutes of arc, we have to multiply by 3438:

d = 3438'*sqrt(2*h/R).

The value of the Earth's true radius is a fixed number. If h is in feet, then we should express the radius in feet. That's about 20,909,000 so now

d = 3438'*sqrt(2*h/20909000).

Pull the constants out of the square root and you get

d = 1.06'*sqrt(h), when h is in feet.

This geometric results was worked out centuries ago, but when dip was measured, the results were always smaller. The constant derived from this simple geometric analysis is too big by about 10%. And the distance to the horizon is correspondingly too small by about 10% (they no longer match). The "square root of height" variation in altitude matched observations and there's no significant debate on that, but only if 1.06' is replaced, on average by 0.97' as the constant of proportionality. It took no great science to realize that this was due to refraction, but there was some "academic" disagreement 200 years ago on the exact value of the refracted constant. Should it be 0.97 or 0.98 or even 0.90? There were cases of editors of navigation manuals, including Bowditch, claiiming that their dip table was better because they used a more "scientific" value for this constant. Those claims were hot air. Today, it's much easier to recognize that the constant just isn't a constant. It's relatively stable most of the time, and the value 0.97 is an excellent choice for an average, but there can be substantial variations. In simple modelled cases, like linear lapse rate variations, different "constants" can be calculated for different weather conditions. If we change the lapse rate of the lower atmosphere (and we know from observations that it is quite variable), we get a different constant. Unfortunately, this simple variation is only a piece of the puzzle. Developing a modified formula for dip that would incorporate real variations in conditions that could be used in the real world is as complicated as predicting fine details of the weather.Here's the message with the dip "cartoon":

http://fer3.com/arc/m2.aspx/Refraction-dip-diagram-FrankReed-mar-2013-g22963-FER

PS: Bruce, consider using "sqrt(h)" for square root of h rather than "sq rt h". The former is standard.

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