[Proj] How to convert a sphere to ellipsoid with correct datum?

[Jan Hartmann]

  Thanks Mikkael, I'll follow that road too. I corresponded over this 
with a Dutch geodesist who programmed a datum conversion tool (Jan 
Hendrikse, http://members.home.nl/hendrikse/), but the results were not 
as good as transforming the triangulation points using the original PROJ 
formula, and afterwards rubbersheeting them to their exact modern 
position. There was a problem wit GDAL, though: it's not possible to 
rubbersheet an already georeferenced map, due to a limitation in the 
Geotiff format. I think now that it can be done by using the VRT format.

And there remains the question of proof. I can (and will) compute the 
datum shift, and can get a map that is within 10m accuracy, about the 
theoretical maximum, but is a datum shift really the reason for the 
deviation? Did they really use an ellipsoid with a different center and 
location in 1850? I find it hard to believe, at least I never found an 
indication for that, and I read the handbooks used in 1850. And Cliff 
Mugnier doesn't think it either 
(http://lists.maptools.org/pipermail/proj/2008-November/004042.html).

So, while I know now how to solve the problem computationally  and get a 
map that is as exact as can be (thanks to all you input, thanks!), I am 
still wondering about the reasons for the deviation. There could be some 
legal issues (although I am not much afraid of those, it's too long 
ago), but for the most I don't wont to be right for the wrong reasons. 
I'm not a great believer in statistical "proofs" where the underlying 
law or model is unknown.

Thanks again for all your responses,

Jan

On 09/07/10 09:04, Mikael Rittri wrote:
> 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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