 Check out the bookstore at IRBS.com |
|
|
|
|||||
|
||||||
From: Bill (no email)
Date: Wed Jan 11 2006 - 16:14:36 EST
> Bill you wrote:
> "I have done three trial runs with the new values, and they seem to approach
> reality with the average of mean refraction values derived from the two
> methods above, a least when height of eye is 0."
>
> Can you elaborate on this 'reality' you speak of? <g> Seriously, how are you
> assessing the results.
>
> -FER
> 42.0N 87.7W, or 41.4N 72.1W.
> www.HistoricalAtlas.com/lunars
Sorry Frank, I keep forgetting that there is very little objective reality
;-) Perhaps poorly stated, my intent was that the results from both the m^2
formula and your method come within spitting distance of each other for a
mean value, and both are almost twice what Bowditch would predict. With the
advance in science I would assume we have information not available to
Bowditch.
As it easier for me to work in feet of rise for my trials, I prefer the
formula that spits out feet directly. Modifying from the site you pointed
out: nautical miles squared * 0.287= lift in feet. I did note that if I
change the constant to 0.265, the results will come within a foot or better
of your 0.15' per mile/minute guideline as tested calculated from 10 to 100
nm. It is relatively easy to convert the constant for values other than
0.15' if one wishes.
I am still working through your later emails. In regards to:
"Well, I wouldn't say it that way. The "problem", such as it is, is that
the tables in Bowditch are calculated for a specific value of the
terrestrial refraction --around 0.16 or so."
Referencing your R/(1-x) it would appear the x value in T15 is 0.168. My
calculations, *assuming* that your and the web site's methods come close to
mean refraction as we currently understand it, would be close to x = 0.312.
I'm not dogging Nate, it is just that times have changed, and as pointed
out, impact traditional navigation.
Back to the beach shots. Even though the adjusted T15 constants get us
closer to the 23 nm target (Sears) and 23.5 nm (Hancock) with height of eye
15 ft and assumed base above water level of 30 ft, both calculations leave
us short. Meaning the observed angles were too high by approx 0.8'. How
much of this difference be attributed to possible thermal inversion?
Bill
|