Nautical studies

Please advise formula for calculating rising/dipping ranges if tables are not available or the height of light is beyond the scope of the tables.

ie Ht of light 155m/ ht of eye 3m

Thanks

Mike

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  1. If you assume a spherical Earth, and if you assume a true horizon without refraction, then the problem boils down to what is the horizon distance for each of the altitudes (eye and light). The maximum range is when the horizons for each is tangent, ie,the horizon range for the eye adds to the horizon range for the light.
    Here is the development for horizon range for a satellite. If you add a second point, the eye, and put the horizon circles tangent, you have the eye-light range.

    See equations 9.2 and 9.8.

    http://en.wikipedia.org/wiki/Belfast_Peace_Lines

    So, for the eye to light, range is given by

    range=eyehorizon + lighthorizon
    =Re/tanRhoeye + Re/tanRoelight

    Where the angles are defined by

    sinRho=Re/(Re+H) equation 9.2

    Now, the radius of earth as a variation with latitude, for this I would ignore it, as the assumption of no refraction will cause greater error. Refraction bends the light around the surface, giving greater range. Refraction is largely unpredictable, due to humidity and temperature gradients above the Sea Surface. The greatest refraction I have seen was in springtime, in the Carribean. This is in my mind the cause of the legendary "ghost ships"appearing and disappearing. If you want to include Re as a variation of latitude, it is easily done with the Cosine function of latitude (but it is just clutter in the calculations).

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    bobpursley
  2. Opps, wrong link. Sorry.

    http://books.google.com/books?id=HfnGxKhcQWUC&pg=PA418&lpg=PA418&dq=spherical+geometry+horizon&source=web&ots=lFC48ZfxZg&sig=iRpF0g3nZCxsSilom9iZIh2X5mc

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    bobpursley
  3. Thanks Bob,

    Got it sorted.

    2.08 x Sq root of the height gives horizon in NM.

    Answer in my question is 29.5M

    Mike

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