--- In
BoundaryPoint@yahoogroups.com, "Ron McConnell"
<rcmcc@e...> wrote:
>
> In working on a question our Dave Patton
> posted on the comp.info.gis and sci.geo.cartography
> netnews groups about the US-Canada border monuments
> # 194 and # 195 and a plaque, and whether the plaque
> is on or south of the border as defined by
> the line of sight between monuments,
> I discovered that while my GCGC great circle, etc.
> program takes in many digits (often more than
> justified by the accuracy of the data)
> and used them for calculations,
> for some long forgotten reason, I had
> limited the digits on output to only 6 good ones.
> When getting down to milliseconds of latitude
> and longitude and inches and centimeters of distance
> this gets important.
> The GCB program didn't have the problem.
>
> The plaque seems to be about 7.47cm or 2.94in
> south of the border, subject to correction.
well first
do you mean the plaque itself
& if so then what point upon it
or do you mean the coords expressed on the plaque
that refer to the center of the 1909 marker
& could you show more details of your computations
i dont mind being 9 to 21 inches or more off
as well as several decimal places too coarse
but i would like to understand why & how i went wrong in both
instances
& please look for one more insert just below
> I'm sure about the .47cm part, just not the 7. part. :)
>
> I have posted an update of GCGC on my website,
> mainly for such close distance uses of BoundaryPoint folks.
> Amateur radio operators for whom GCGC was
> originally intended work in much longer distances
> like 100s or 1000s of kilometers or miles.
>
> [from sci.geo.satellite-nav netnews group]
>
> "I am a surveyor and use differential GPS every day.
> I can achieve accuracies of less than 20mm
> in xyz in clear areas." - Mark xxx@n...
>
> From Sigurd Humerfelt's web site notes on
> "The Earth According to WGS 84"
> "Every degree of Latitude"
>
> <http://home.online.no/~sigurdhu/Grid_1deg.htm>
>
> N/S 49° (theoretical US-CA border)
> 1 min lat = 1853.50 m 1 min long= 1219.53 m
> 1 sec lat = 30.8917 m 1 sec long = 20.3255 m
>
> Then at N/S 49º
> 20mm Lat = 0.02m / 30.8917m/" = 0.0006474" Lat
>
> 20mm Long = 0.02m / 20.3255m/" = 0.0009840" Long
>
> Or better than 0.001" of Lat and Long with DGPS
>
> This is consistent with "altheiak's" MDVADC
right
up to a point
tho you must mean my mdvawv data
where they are actually given by the surveyor in 5 decimal places
of degminsec
end inserts
thanx
cheers
& lets see some more of this magic
> surveyor's lat/long data to 0.001 seconds.
> Be skeptical about anyone who claims better than that
> or puts more into or out of a computer program.
>
> Cheers, 73,
>
> Ron McC.
> w2iol@a...
>
> Ronald C. McConnell, PhD
>
> WGS-84: N 40º 46' 57.6" +/-0.1"
> W 74º 41' 22.1" +/-0.1"
> FN20ps.77GU31 +/-
> V +5058.3438 H +1504.2531
>
> http://home.earthlink.net/~rcmcc
>
> "Any sufficiently advanced science
> is indistinguishable from magic."
> - Arthur C. Clarke