# NavList:

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

**Re: Summary of Bowditch Table 15**

**From:**Bill B

**Date:**2005 Jan 26, 17:12 -0500

> Doh! So what is "D" then, and how does one use the table? D is distance in nautical miles from the observer to the object. Top or bottom? What does it matter? In geometry there would be a slight difference, in real life other variables will more than wipe out any minor differences. As to how to use the formula, simple. Shoot angle between the visible horizon and the top of the object. Correct for index error and subtract dip for observers height of eye. Use that angle as the entering argument in "Angle" column on the left or right of the table. Subtract height of eye (ft) from the height above water (ft) of the object. Use this figure as the entering argument in the "Difference between..." row at the top of the table. Use the figure where the angle row and difference column meet as your final answer (in nautical miles). No need to calculate your horizon. Either you can see the base of the object (use table 16) or you cannot (use table 15) ====================================================================== As I perceive it, using plane geometry if I could see the whole object, I can use the angle and object's height (opposite side) to determine the adjacent side in feet (Tan = opposite over adjacent). But I cannot see the whole object in this case because the Earth's curvature puts a "hump" between my eye and the object. The further I am from the object, the bigger the hump so the less I can see. So if I know the angle I would get if I could use my x-ray vision to see through the water, the angle I can see, have a formula for the affect of the hump as distance varies, I can through (the magic of math) determine my distance from the object. How the formula works I leave to the math whizzes. Clues are we see the square root of H-h/0.7349, which would equal H-h * 1.36. Pretty close to geographical range and horizon constant of 1.169 or 1.17. We also see tangents of the angle, and they are divided by 0.0002419. The same as multiplying by approx. 4134. That number looks suspiciously close to the Earth's radius. We also play with the affect of the "hump" expressed as an angle by subtracting dip. But there are no old bold empiricists, so I'll quit there ;-) Bill > > My height of eye is 30 ft. > My horizon is 6.4 NM away. > (Distance to Horizon for h in feet = 1.17 * sqrt(h)). > And my Dip is - 5.31'. > (Dip in minutes of arc for h in feet = -0.97 * sqrt(h)). > The lighthouse over the horizon is 100 ft high from its sea level. > > I measure a vertical angle of 00d 05.40' with my sextant (not that I can > measure to a precision of hundredth of minute of arc). > The corrected vertical angle is 5.40' - 5.31' = 0.09'. > (H-h) = 100-30 = 70 ft. > From the table then, D = 3.8 Miles. > > D is, I thought, the distance from the sea horizon to the top of the object. > I presumed that I should add my distance to my horizon (6.4 NM) and the > distance from the horizon to the top of the object (3.8 NM) to get a total > distance of 10.2 NM, with a 0slight vertical dogleg two thirds of the way > along. Is that incorrect? > > As a rough check I know that: > My horizon is 6.4 NM away from an eye height of 30 ft. > The top of 100 ft lightouse sees a horizon 11.7 NM away. > So when I can just see the top of the lighthouse at my visible horizon, we > are separated by about 18.1 NM. > At least the magnitude seems reasonable. The lighthouse must be closer than > 18.1 NM when I can measure a greater vertical angle than 00.00'.