WGS84 to metres conversion (very accurate)

Hello,

I was wondering if anyone has a very accurate formula to calculate
meters to lat/lon (wgs84) and visa versa. The accuration has to be
around centimeters. The formulas I found on the web did not have
that amount of accuration.
I am very aware of the fact that GPS which gives the coordinates has a
deviation of 1-2 meters, but that isn't of any problem for my specific
research.

Thanks.

 

See the coordinate transfer soltware and online links i the
box near the top of
http://edu-observatory.org/maps/waypoints.html

such as NGS Geodetic Tool Kit
http://www.ngs.noaa.gov/TOOLS/

 

snip

You posted links to software and even more links. I need formula's

 

For UTM, NGA TM 8358.2, from
<URL:http://earth-info.nga.mil/GandG/coordsys/csat_pubs.htm>,
has formulas that give centimeter accuracy.

For UTM and other projections,
Snyder, John P. _Map Projections--A Working Manual_.
U. S. Geological Survey Professional Paper 1395.
(available in hardcopy only)

Or do you mean an Earth-centered Cartesian coordinate frame with
coordinates in meters?

 

The idea is that i have the WGS84 coordinates of some Access Point
(WiFi). Now I need the coordinates for when I am say 3 meters to
the left. My specific task is to determine that position as precise
as possible.

 

Aviation [Navigation] Formulary V1.37 by Ed Williams
http://williams.best.vwh.net/avform.htm

 

The formula described here might do the job:

<URL:http://williams.best.vwh.net/avform.htm#LL>

It's a great-circle approximation, but for distances on the order of 3
meters, the error should easily be within your centimeter threshold.
The advantage is that the formula is much simpler than the UTM
transformation.

Note that
(1) d in the formula is an angle, not a linear distance; read
the introduction section at
<URL:http://williams.best.vwh.net/avform.htm#Intro>
(2) that document uses the convention that west longitudes
are positive. If you find other formulas in that document
useful, you may need to take that convention into account.

If you don't think either that solution or one using a projection like
UTM will be adequate, you might also be interested in the coordinate
transformations described here (geodetic to EFG to a local NED/ENU and
vice-versa):

<URL:http://www.eiserloh.org/~peter/doc/tspidoc/html/main.html> or
<URL:http://www.eiserloh.org/~peter/doc/tspidoc/tspidoc-0.7.18.pdf>

 

You don't need fancy algorithms to obtain millimeter or better accuracy at
distances of a few km or less. See http://groups-
beta.google.com/group/sci.geo.satellite-nav/msg/0bfca0bf8a986395

 

http://williams.best.vwh.net/avform.htm#flat

will be good to about 1 part in 10^12 at three meters.

Ed

 

Mathematicians should not use GPS. They should leave it to walkers,
mountaineers, drivers and sailors.

 

I was wondering if anyone has a very accurate formula to calculateI must admit that I have my doubts about any real application requiring
centimeter accuracy. Such accuracy can only ever be theoretical and of
little practical use. Unfortunately (?) we do not live on a perfect shiny
sphere (or even spheroid). Our planet is covered in lumps and bumps and
dips etc. etc. WGS84, like all datums is just an approximation which works
well in some places, not so well in others. I fear that centimeter accuracy
in real-world surface distance measurement may be asking too much of it -
but maybe that's not your intention.

Perhaps a related question should be centimeter accuracy in what distance?
If you want distances of the order of thousands of kilometers to the nearest
centimeter then surely that is expecting too much. I have a suspicion that
even centimeter accuracy in distance of the order of a few meters might be
expecting too much but I will defer that to those with better technical
knowledge of the issues.

Keith