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Thank you all for your responses to my initial question. I will try to
summarize where we have got to.

1: When I complained that Table 9/15 doesn't work as advertised, I was
thinking of dip as a minor correction (in the sense that refraction is)
and hence that, after correction for dip, an observer should still see
more of a distant object as he increases his height of eye. As George
has pointed out, the reason that objects appear to rise over the horizon
as height of eye increases is that dip increases. Correcting for dip
changes the whole thing.

Specifically, correcting for dip changes the reference plane from the
apparent horizon to the sensible horizon. As height of eye increases,
the sensible horizon obviously moves further from the centre of the
Earth, which means that it cuts the masts of a distant ship closer to
the truck. Hence, the angle subtended at the observer's eye by the bit
of mast projecting above the sensible horizon is reduced as height of
eye increases.

That is what Table 9/15 shows and so I have no reason to question the
meaning of the table.

[Jim: Does that explanation make sense to you? In your last posting, you
said that you still thought the direction of change in the angles was
wrong in the Table. I previously thought that but George has, I think,
explained the confusion.]

2: Specifically, I withdraw my suggestion that it shows the distance
from the observer's horizon to the distant object. My reply to Nels
should have disposed of that one anyway.

3: However, I cannot confirm numerically that Table 9/15 is correct.

Consider a 100ft mast seen from 5 miles away by an observer with 10ft
height of eye. The 90ft of the mast extending above his sensible horizon
should subtend an angle of about 10' at his eye, yet entering the table
with (H-h)=90ft and angle 10' gives a range of 4.3 miles.

Let the observer climb a further 60ft up his own mast and the remaining
30ft of the observed mast showing above the sensible horizon should
subtend about 3.4'. However, Table 9/15 would then reduce the range to
3.3 miles.

I suspect that these differences can be accounted for by refraction but
the way to be sure of that is to have someone derive the Bowditch
formula as a formula for range from observer to object, based on the
height of the object above the sensible horizon. Maybe one of the
mathematically-gifted list members could do that. I cannot.

4: Any such derivation requires, of course, that we can agree on what
the Bowditch formula is.

George has reported that the 1981 Bowditch makes it:

d = sqr{(tan A / .0002419 )^2 + ((H-h) / .7349) - (tan A / .002419)}

My 1995 Bowditch has seen the attention of some over-zealous editors,
who "corrected" that to:

d = sqr{(tan A / .0002419 )^2} + ((H-h) / .7349) - (tan A / .0002419)

(Note the extra "0" as well as the shift ion "}".) Jim says that the
2002 edition has the same.

In contrast, Bill's numerical analysis has shown that the only version
which makes sense is:

d = sqr{(tan A / .0002419 )^2 + ((H-h) / .7349)} - (tan A / .0002419)

So, Bowditch has had this wrong for at least 25 years and still has it
wrong. I wonder when it was last correctly quoted. [I'm guessing but I
suspect the actions of generations of editors since the last attentions
of a compiler who actually thought about the meaning of the material,
instead of just carrying forward the last edition, with amendments.]

Jim has pointed out that this issue was discussed on the list
previously, when George correctly deduced what the formula should be. (I
don't feel so bad about having forgotten that exchange, since George
evidently did too!)

5. That leaves, by my count, only two outstanding yarns within this
thread (to stretch a metaphor): Jim asked:

"How should the Dip correction be applied?  Add?  Subtract?"

It should clearly be subtracted from H[s], just as with a celestial sight.

"See my new Figure 1 and tentative explanation of the Dip correction for
Table 15 at
http://jimthompson.net/boating/CelestialNav/CelestNotes/Bowditch15.htm"

Jim: I think you have complicated things by drawing a distant object
which doesn't rise above the observer's sensible horizon. For such a
case, you have the angles correctly labelled.

However, your numerical example (as is true of most of the numbers in
Table 9/15) is for an object which does rise above it. In trying to
reconcile those two different cases, you have put a double negative into
your numerical example. Take that out and draw yourself a new, higher
object, extending above the sensible horizon, and all should fall into
place.

"How is H-h calculated?  I had assumed that one subtracted the
observer's height of eye from the known height of the object above its
sea level"

Yes.



Trevor Kenchington



--
Trevor J. Kenchington PhD                         Gadus@iStar.ca
Gadus Associates,                                 Office(902) 889-9250
R.R.#1, Musquodoboit Harbour,                     Fax   (902) 889-9251
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