[Proj] How to convert a sphere to ellipsoid with correct datum?
Mikael Rittri
Mikael.Rittri at carmenta.com
Tue Sep 7 02:04:17 EST 2010
Janne wrote:
> There are methods to find those parameters when given some
> data to be fixed. 60 m should not be any problem.
I agree with Janne. In fact, I think you, Jan, were right
in your earlier thread, that started with
http://lists.maptools.org/pipermail/proj/2008-November/004041.html
that what you need is a datum shift from a Dutch datum of
the 1850s to the more modern Amersfoort datum. Or, if you
want to use cs2cs, you need a datum shift from the 1850s
datum directly to WGS84.
Your problem is just that you haven't found a published datum shift.
In the November 2008 thread, a few people said that a north-south
mismatch of about 650 meters was too large to be blamed on differences
between datums. They seem to have convinced you that it is not a datum
shift you are looking for. However, I think the 650 meters mismatch was
artificially created by your attempt to do an ellipsoid-to-ellipsoid
conversion without any datum shifts, which is the same as implicitly
assuming that the Ackermann ellipsoid of the 1850s datum and the Bessel
ellipsoid of the Amersfoort datum have coinciding centers and axes.
(As Frank also noted, http://lists.maptools.org/pipermail/proj/2008-November/004050.html ).
This implicit assumption is just wrong. I'd guess that it is 600 - 700
meters between the ellipsoid centers.
If you, instead, had just directly compared longitudes and latitudes
between coordinates in the 1850s datum and in Amersfoort, regarding
1 minute of latitude as 1 nautical mile, I think you'll find that
the north-south mismatch is more like 50 - 60 meters. This means
that it is an ordinary datum shift you are looking for.
What I'd do, if I were you, is to take those positions that you know
in both datums (the 1850s one and Amersfoort). Those in Amersfoort I
would convert to WGS84 by the appropriate 7-parameter datum shift.
Then, I would convert the coordinates to geocentric Cartesian X,Y,Z
coordinates, both for the 1850s datum and for WGS84. Now, for each
position, take the vector difference from its X,Y,Z-coordinates in
the 1850s datum to its X,Y-Z-coordinates in WGS84. Then take the
average of all those difference vectors. This average should be
a reasonable 3-parameter datum shift from the 1850s datum to WGS84.
The standard deviation of the difference vectors, and the outliers
compared to the average, should give you an estimate of the accuracy
of the 3-parameter datum shift.
If you have enough common points, you can also derive a 7-parameter
datum shift, which could compensate overall rotation and overall
scale difference between the 1850s datum and WGS84. But I don't
really know the mathematics for deriving a 7-parameter datum shift.
There should be software available, but I don't know if there is
any open source or free software.
If the 1850s datum has many local distortions, a 7-parameter datum
shift for all of the Netherlands might not be so accurate anyway.
Another approach would then be to construct many local 3-parameter
datum shifts, each one based only on the common positions in a
local area, like a county, or perhaps the common positions that
are nearest each of your local church of interest.
Best regards,
Mikael Rittri
Carmenta AB
Sweden
www.carmenta.com
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