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John, you wrote:
"Ahhhh Frank, see what I mean----"

Not really. Could *you* explain what you mean for me? See, as I have read 
these posts, I see people trying to be careful with terminology, and 
definitely also people get confused in their intial celestial education by 
the fact that different AP's are chosen for different bodies in the same 
round of sights with certain tabular algorithms. And that aspect IS confusing 
for most beginners. But it's not because of the terminology, as far as I can 
see (so far!).

And you wrote:
"All these posts, and look at the confusion:  The DR is an AP, it's one
particular choice of an AP.  And tables have nothing to do with it."

John, it seems to me that this is pure semantics --differences in the way 
people define the "assumed position". If we define the concept of "assumed 
position" narrowly (as a point where we are not located chosen only for 
convenience of solution) then the DR is not an AP. If we define it broadly as 
the point from which any intercept is calculated in the St Hilaire method, 
then a method plotting from the DR uses the DR as an AP. Under that 
definition, it's the same thing. The interesting thing here is that the 
difference in the way one defines it, narrowly or broadly, can impact 
communication with people who have *already* learned celestial navigation in 
different ways (for example, you talking with Peter Fogg in these messages), 
but it's not a problem for people learning celestial navigation from one 
source or one instructor (since in the latter case, you will encounter only 
the narrow or the broad definition, not both). But maybe in the end, this is 
the very confusion you're talking about. Is it? 

And you asked:
"Has anyone answered the question WHY we use an AP in the intercept method??  
To phrase it another way, since we know everything needed to calculate the 
celestial LOP (the GP's location and the altitude), why not just calculate 
the LOP directly?"

Well, you can do that today. Historically it was prohibitively inefficient in 
terms of computation cost. But today if I observe Venus 4d 12.5' above the 
horizon, I can easily calculate every spot on the Earth where this would be 
the observed altitude. I wouldn't recommend that approach even today on a 
computer. With normal sights, there would be no value in that direct 
calculation. I'm not entirely sure what you're getting at here, so it's quite 
possible I completely misinterpreted your question. Do let me know if that's 
the case! ;-)

-FER



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