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    Re: Measuring (and Calculating) Dip
    From: Greg Rudzinski
    Date: 2013 Mar 3, 04:40 -0800

    Brad,

    If the observed horizon is taken at the wave peaks then height of eye should also be taken from the average wave peaks from the location of the observer when aboard ship. If at shore then a correction would have to be made to compensate for horizon waves. Perhaps this is what you are getting at. Another factor to look at with less significance is the difference in Earth's radius from pole to equator (equatorial bulge as my astronomy teacher said pointing to his well developed belly girth ;-)

    Greg Rudzinski

    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
    Brad


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