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Jim Thompson wrote, about Bowditch Table (9 or 15)

>Getting the sensible horizon below the top of the object is key to
>understanding this table.  If the sensible horizon shoots into space above
>the top of the object, then the relationships in Table 15 are "wrong".  In
>real life the sensible horizon is unlikely to be above the object, I
>presume.

No, not wrong. Indeed the first page of my Table 9 specifically caters for
negative angles of up to 4 arc-minutes. It's quite possible to see a
distant object at a small negative angle, below the true horizontal,
limited only by the dip, which is also below the true horizontal.

>The "correct" (consistent with Table 15) relationships are these, for an
>object with constant height, stationary distance between ship and object
>over the horizon, and decreasing height of eye:
>
>- Measured vertical angle decreases (observer sees less of the lighthouse).
>- Dip (angle between sensible and visible horizons) decreases (sea horizon
>gets closer to observer).
>- Corrected vertical angle increases (consistent with the table, I know see,
>but not intuitive).

OK so far

>- D increases, if D is the distance from the observer's horizon to the
>object.

No. From table 9, as the height-difference H-h increases as the observer
descends his mast, then the increasing vertical angle corresponds to the
same distance d as before, because that hasn't changed. The distance d is
NOT the distance from the observer's horizon to the object. That was
someone's misunderstanding; Trevor's, perhaps, that has since been
recanted. d is the distance between the observer and the object.

>Doh!  So what is "D" then, and how does one use the table?
>
>My height of eye is 30 ft.
>My horizon is 6.4 NM away.

You don't need to know that.

>(Distance to Horizon for h in feet = 1.17 * sqrt(h)).
>And my Dip is - 5.31'.

Yes, you do need to know that.

>(Dip in minutes of arc for h in feet = -0.97 * sqrt(h)).

Well, I would think of the dip as +0.97 * sqrt(h) and subtract it, but
taking the dip to be negative, and then adding it, works just as well.

>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.

No, look again. That's the value the table gives for a corrected angle of 0
deg 09', not 0 deg 00.09'.

Jim's observation has resulted in a value of one hundredth of his presumed
amount, 0 deg 00.09'. As he says, he can't measure to that precision. Or
estimate the dip to that precision, either. But let's go along with those
numbers.

The nearest entry in the table for Jim's 0deg 00.09' is that for 0deg 00',
which gives 9.7 miles as the resulting distance of the lighthouse from his
vessel.
>
>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?

All quite incorrect. The 9.7 miles answer was the solution to the problem.

>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'.

I hope it's clearer now.

George.

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