# NavList:

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

**Re: Bowditch (2002) Table 17**

**From:**Bill B

**Date:**2012 May 01, 03:29 -0400

As the math, engineering and physics gurus have passed on your question to date, I'll attempt a layperson's explanation. Once I build a straw man, I'll wager they chime to correct my blunders :-) (It's been a long time since I reverse engineered the equations). To over simplify, Table 15 is a case of measuring the portion of the object still visible and adjusting for the unseen portion below the horizon due to curvature of the earth, then adding to the observed angle for refraction. Way way over simplifying, Table 17 is similar to measuring the height of an nearby object above ground level. I surmise it then elegantly uses height of eye to divide it into to right triangles. (Table 16 says that directly.) Throw in a few tweaks for curvature of the earth and refraction... In table 15 you enter height of object minus eye and and the angle (height) of the exposed portion of an object of known height over the horizon. Similar to taking an angle from a celestial body, dip (height of eye) comes into play. Refraction also comes into play. As does the curvature of the earth. If memory serves,there is indeed an editorial error in the equation given in 2002. (H-h)/0.7349 should be included under the square root symbol. As a side bar, Frank Reed did some observations of Chicago, IL skyscrapers from a beach in Indiana. We were both of the opinion the Bowditch constant of 0.0002419 is a bit heavy. Frank suggested an alternative and I derived my own. Both were close to spot on 17 miles off Chicago when I tested them from a sailing vessel. I cannot say that of the Bowditch constants. If you are interested I can dig into my archives and find our proposed constants. Table 17 is a bit more up close and personal. (Although on page 560 there is definitely a typo, the table does explain it correctly.) Given the same height of eye, the distance to the horizon in nm is 1.169 * square root of height of eye (in feet). In table 15 the base of the object is below the horizon so a good distance away. In table 17 the base is visible and the horizon is above the base. We still have curvature of the earth and refraction to deal with. I understand I have not fully answered your questions. As the dimmest bulb on the NavList marquee, it's my best shot. Bill B On 4/30/2012 4:46 PM, slk1000@aol.com wrote: > It has been a while since we discussed distances by vertical angles > (2005 and 2009). I just ran into something I hope someone can clear up > for me. In the Explanation of Tables on page 560 of Bowditch 2002, it > says that for Table 15 the sextant angle is corrected for index error > and dip, but for Table 17 it is only corrected for index error. Why no > dip if height of eye is part of the formula?