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    Re: Measuring (and Calculating) Dip
    From: Brad Morris
    Date: 2013 Mar 20, 12:43 -0400
    Hi Bill
    I applied the dip equation of 0.02977*sqrt(h) of the so called 'standard dip model' to your data. 
    Height Of Eye, Face Left, Face Right, Avg, Computed
    1.89 2' 48" 2' 11" 2'29.5" 2'27"
    2.14 3' 05" 2' 35" 2'50"    2'36"
    3.10 3' 17" 3' 13" 3'15"    3'09"
    3.92 3' 27" 3' 36" 3'31.5" 3'32"
    14.5 8' 13' 7'  39" 7'56"    6'48"
    Therein lies confirmation that the theodolite will be more than capable of dip measurements.  Excluding the case where the height of eye came from elsewhere, your minimum excursion was 1/2" and your maximum excursion was 14".  Nice!
    You wrote:
    sometimes the horizon appears to rise and fall by up to 30 seconds
    While Frank is going to cringe, I will point out that if you are seeing the peak and the trough of a one meter wave at the horizon, then the (bullshit) equation produces a result of 27 arc seconds.  Coincidence?  Perhaps, perhaps not!  Can you remember the period over which this rise and fall occurs?  Knowing the period will allow us to associate the rise and fall with waves, or not! 
    I also note some tsunami buoys in your area which show a 1 meter rise and fall to the column height of the water.  They are Buoys 55015, 55013 and 51425.  These buoys are not producing the wave information that I am used to but the 1 meter variation is crisp and reproducible.  I
    You wrote
     with high magnification the horizon is irregular and difficult to define
    Most folks view waves like ripples in a pond, like big circles.  In fact, they have a length along the crest of the wave.  That is, not big circles, just a short line.  Put enough short lines in a wave field, and you will have a jagged appearance.  From memory, I would like to say waves have a length along the crest which is roughly 7 times the height.  Please do note that this is NOT a peak to peak measurement of wavelength and its associated wave period.  This is orthogonal to that, along the peak.  This will result in a jagged appearance at high magnification, as individual waves will indeed be included within the Field of View of your optics.  So for your 1 meter wave, it will be roughly 7 meters along the crest. 

    On Wed, Mar 20, 2013 at 2:10 AM, Bill Morris <engineer@clear.net.nz> wrote:

    Here are some data from last Tuesday at Henderson Bay in 34deg 44'S, 173 deg 07'E.
    Cloudless sky. Air Temp 25 C; sub surface temp 21 C. Wind speed zero. Wave height negligible though swell of about 1 metre visble off-shore (perhaps ex Typhoon Sandra).
    Height of eye in metres is given first, followed by dip face left and face right.
    1.89 2' 48" 2' 11"
    2.14 3' 05" 2' 35"
    3.10 3' 17" 3' 13"
    3.92 3' 27" 3' 36"
    14.5 8' 13' 7' 39"

    The last height of eye was taken from Google Earth. The others were measured with staff and level. AS I have noted previously, sometimes the horizon appears to rise and fall by up to 30 seconds and with high magnification the horizon is irregular and difficult to define.

    For those who are not familiar with it, I have attached a diagram showing the light path of the Soviet N5 Dip Meter referred to in Brad's post and a couple of views of the instrument. It is a compact and light instrument. It is a little difficult to use because the movements required to view both horizons and bring them into coincidence are counter-intuitive, but once the tricks are mastered it gives the dip directly with a precision of 0.2 minutes on an easy-to-read scale.

    Bill Morris
    New Zealand

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