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Re: Measuring (and Calculating) Dip
From: Marcel Tschudin
Date: 2013 Mar 3, 16:48 +0200

The lack of any statistical information on how the measurements are expected to scatter around the calculated one is the drawback of the two publications and also of your calculation. We do not have a mean to qualify your difference between measurement and calculation which in terms of navigational accuracy could just be a lucky shot. At the time when the measurements mentioned in the two publications were performed it must have been a very tedious task to analyse large numbers of data. These days this can be done much easier, often requiring only a few mouse clicks.

Due to the probably considerable scattering differences between observation and calculation and due to the annual cycle for the temperature difference a statistical analysis should probably contain a few hundred measurements collected possibly also over a few years in order to arrive at conclusions which are "somehow" representative. This should however not prevent you from looking at intermediate analyses from smaller data sets (like e.g. the first shots you made now) but it always requires recognising their limitations.

If you intend to continue to take such measurements then I recommend you to save the data from all observations in a spreadsheet for future analyses.

Marcel

On Sun, Mar 3, 2013 at 6:33 AM, Brad Morris wrote:

Hello Marcel

Thank you for the clarification.

This makes the dip formula
dip = 1.74*sqrt(h) - 0.47*(AirTemp - WaterTemp)
where dip is in arcminutes
h is the height of eye in meters
AirTemp is the temperature of air at eye level, degrees C
For this I use the buoy air temperature
WaterTemp is the water surface temperature, degrees C
For this I use the buoy water temperature

Next, we must consider the effect of waves. When observing for dip (or using the horizon as a reference for an altitude) you are actually using the crests of the waves at the horizon, not the mean sea level at the horizon. The above equation for dip is for the mean sea level.

The angular height of waves at your horizon is
wvhtCorr'n = atan((wvht/2)/(3860*sqrt(h)))*60
where wvhtCorr'n is the waveheight correction, in minutes
wvht is the measure of the larger of wind waves or swell at the buoy
which is a peak to trough measurement, we want mean sea level
to peak. Meters
h height of eye, meters
You may recognize the term 3860*sqrt(h) as this is the distance to the visible horizon. So we are taking the arc tangent of Y over X, which is 1/2 the peak to trough wave height divided by the distance to that wave height! Simple Trig!

Let Height of Eye h=3.268 meters. I was standing on land, and I measured the vertical distance to the water.

From 44039 (Long Island Sound) we have wvht=0.1 meters, air=3.5 deg C and water=2.8 deg C.
Dip = 1.74*sqrt(3.268)-0.47*(3.5-2.8)
= 2.82 arc minutes
wvhtCorr'n = atan((0.1/2) / (3860 *sqrt (3.268))) * 60
= 0.02 minutes (0'1")

From 44017 (Atlantic Ocean, near Montauk Point) we have wvht=2.3, air 5.4 deg C and water 5.6 deg C
Dip = 1.74*sqrt(3.268)-.47*(5.4-5.6)
= 3.24 arc minutes
wvhtCorr'n = atan((2.3/2) / (3860* sqrt(3.268))) * 60
= 0.57 arc minutes (0'34")

So my total dip (calculated) is 2.82 - 0.02 + 3.24 - 0.57 = 5.48 minutes (5'29")

My measured dip was 4 minutes 55 seconds (4'55").

Therefore, the total error in calculation was (5'29" - 4'55"=)+34", but since this is two dips, we must divide by two for an average dip error of +17". Just as a side note, the winds were calm at the time of measurement. This is evidenced by the 0.1 meter wave height on the Long Island Sound (3.9 inches peak to trough!).

CONCLUSIONS:
1) The wave height correction proved to be decisive for this measurement. I agree with Hasse in this regard. Nowhere in the standard dip equation do I see an accomodation for wave height (sea state).
2) The precision wave data afforded by the NDBC Buoys is far better than Friedenleben's attempts at this data. We have precise wave measurements, not Beaufort sea state estimates as a function of wind speed.
3) The Buoy air temp is measured at 4 meters above sea level. Consider this in comparison to the height of my eye at 3.268 meters. It is acceptable to use this as the temp of the air at height of eye, for my measurements.
4) The Buoy water temp is 0.6 meters below the surface. This is to be compared to Friesenleben's boats churning about in the water, dipping buckets into the surface and then measuring the water with a thermometer. What could go wrong with that ;-)

Here's an un-verified theory: To further understand the thermal model, I believe we will have to take into account the wind speed over the water. For a calm day, the lowest band of air will be approximately the water temperature, as the air and water are in contact without motion. For higher wind speeds, I believe the lowest band of air will not take on the water temp, rather it will maintain itself closer to the air temp at 4 meters! For calm days, we have a ramping range of temperatures, from air to water. For windy days, we will have more of a step function.

Here's a plain statement of fact: Friesenleben suffered from poor data, quite uncontrolled in quality. This could clearly have affected his results and conclusions.

Best Regards

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