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    Re: Dip Measured versus Dip Calculated
    From: Brad Morris
    Date: 2013 Apr 7, 13:43 -0700

    Hi Bruce

    I absolutely agree that the delta could be from differing environmental conditions. Indeed, that was one of the possible sources listed.

    As Marcel has indicated several times over, the temperature GRADIENT is much more important than a single temperature. More over, from Dr. Andrew Young, we have a gradient equation

    dip, minutes = 5.04 sqrt(.1123h + t0 -th)
    temp units in degrees C
    h units meters

    I went to NDBC Buoy 44039, to get met data for the air over LI Sound.
    http://www.ndbc.noaa.gov/data/realtime2/44039.txt
    On April 4, 2013
    At 12:30 the air temp was 1.9C and the water was 4.1C
    At 5:30 the air temp was 3.7C and the water temp was 4.3C
    This corresponds to your measurements 5A and 5B of 4.42 moa and 2.99 moa correspondingly.

    Lets solve for t0 so we can see what the gradient looked like ALL OTHER ERRORS ASSUMED ZERO.

    In the first example, we have 4.42'=5.04'*sqrt(.1123*(13.5*.304)+t0-1.9)
    t0 must equal 2.208 degrees C. That is the air temp just at the surface of the water was warmer than the air temp approx. at the height of eye, but colder than the sure]face temp of 4.1C

    In the second example, we have 2.99=5.04*sqrt(.1123*(13.1*.304)+t0-3.7)
    t0 must equal 3.604C. Now the air at the surface is nominally COLDER than the air temp at the HoE and the water temp.

    Brad


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