# NavList:

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

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Re: Waves & Dip
Date: 2017 Apr 23, 13:12 -0400
Mike

I remain unconvinced that the reference you state is correct.  I think you may be swayed by this discussion.

The wavelength for long period swell is shorter than you think.  16 second swell is nominally associated with major storms, like hurricanes.  At 16 seconds, the wavelength  (distance between crests) is roughly 400 meters.  For shorter period swell, in seas that you will likely be yachting in, the wavelength will be shorter.  I'll assume your height of eye is roughly 8 feet, so the horizon is ~3.3 miles (5300 meters) distant. So out at the distant horizon, what you see is one crest slightly obscured by the one in front, as the grazing angle becomes shallower and shallower.  At 3.3 miles, the grazing angle is ~1.5 arc minutes.  Eventually, all you see are the very crests of each wave, until it becomes an indistinguishable blur.  You never see the troughs, only the crests.

But your yacht is also at the crest of the wave.  Under the assumption that the wave field to the horizon contains similar amplitude waves, your yacht has maintained a similar height above the observable horizon as if it was in perfectly calm water.  It's floating at the top of a wave, yet all you see are the tops of waves.   It's roughly the same as a flat sea.  So there is no adjustment to the HoE.

Of course, I'm ignoring the heave of the yacht as it rides up and down the waves.  Similarly, waves aren't perfectly uniform in height, so the horizon won't be a smooth line, rather, it will be a rough jumble.

But estimating the wave amplitude, dividing it by two and adding it to your HoE?   Let's examine that for a moment.  Suppose you were at the crest of a rogue wave, looking down at a flat sea all around you.  Knowing the height of the rogue wave, and your nominal HoE, you would add them together, no division by two; because your eye was raised, relative to the observable horizon.  Suppose you were on shore.  The waves will have "raised" the apparent horizon, but not you.  We would subtract the wave height from the HoE, again no division by two.   But for a yacht at the crest of a wave, taking an observation using the surrounding wave field as a reference?  I think that's a zero adjustment to HoE.

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