Sorry, i lost my mails :-(
Also, i have a solution for the 3d transformation:
NB. common points in the 1st system
crd1=. n 3 $ ...
NB. common points in the 2nd system
crd2=. n 3 $ ...
NB. hub caluclating funtion
hub=. +/ %#
NB. hub of the 1st system
hub1=. hub crd1
NB. hub of the 2nd system
hub2=. hub crd2
NB. 1st system minus their hub
crd1h=. crd1-"1 hub1
NB. 1st system minus their hub
crd2h=. crd2-"1 hub2
NB. rotation matrix from 1st system to 2nd system
r12=. crd2h % crd1h
NB. rotation matrix from 2nd system to 1st system
r21=. crd1h % crd2h
NB. remaining errors in the 1st system
e12=. crd2h-(crd1h +/ .* r12)
NB. remaining errors in the 1st system
e21=. crd1h-(crd2h +/ .* r21)
NB. Sample (m) points in the 1st system
sample1=. m 3 $ ...
NB. Sample points in the 1st system minus hub of the 1st system
sample1h=. sample1-"1 hub1
NB. Sample points in the 2nd system
sample2=. hub2 + (sample1h +/ .* r12)
I have two vectors (hub1, hub2) and a rotation matrix. This is good but
these parameters still not the seven parameter. Can I calculate from
these vectors and rotation matrix the 7 parameter?
--
Zoltan
Mike Day írta:
This is reminiscent of Mr Faludi's request for help on 28/4/05
when he confirmed that he wished to deduce 7 parameters from
data. The transformation was then "Bursa-Wolf" rather than
Helmert; presumably these are important names in Geophysics.
The Bursa-Wolf problem concerned the same 2 sets of 10 xyz
coordinates as for the Helmert query:
wgs84 -: src NB. wgs84 from Bursa-Wolf qn
1
grs67 -: dst NB. grs67 from Bursa-Wolf qn
1
I offered him a Newton-Raphson method solution; it was
apparently not accurate enough for dz. Our correspondence
ceased then, so I don't know whether my method was at fault
or if the comparison tolerance was too low. I'm not
inclined to repeat the exercise, but am happy to forward
my script to anyone who's interested, or you might find it
in my message in the archives, dated 12/May/05 and headed
"Re: [Jforum] Bursa-Wolf 7 parameter"
Kym Farnik also corresponded on the Bursa-Wolf question.
Mike
Devon McCormick wrote:
Yeah - I found this, too, which describes the inverse problem of
deducing
the parameters
from the transformed coordinates:
http:NB. www.maths.dundee.ac.uk/~gawatson/helmertrev.pdf .
On 1/23/07, David Mitchell <[EMAIL PROTECTED]> wrote:
Perhaps this describes the problem: http:NB.
www.killetsoft.de/p_svpa_e.htm
--
David Mitchell
Devon McCormick wrote:
> Are you saying you need to work backwards from the source and
> destination points to derive the 7 parameters?
>
> On 1/22/07, Faludi Zoltán <[EMAIL PROTECTED]> wrote:
>>
>> Hello!
>>
>> How can I implement the calculation of 3D helmert transformation
>> parameters (7 param)?
>> dx: shift in meters
>> dy: shift in meters
>> dz: shift in meters
>> rx: rotation in arc seconds
>> ry: rotation in arc seconds
>> rz: rotation in arc seconds
>> ppm: scaling in parts per million
>>
>> I have two matrices with the common points of the source (src)
and the
>> destionation (dst) coordinate system.
>>
>> for example:
>> src=. 10 3 $ 4190468.039 1419136.810 4578982.785 4198120.436
1411059.537
>> 4574631.898 4198048.482 1417658.260 4572620.884 4192498.898
1425966.119
>> 4575143.392 4186048.597 1425904.183 4580905.806 4184439.769
1410536.684
>> 4587280.046 4186429.631 1417172.136 4583253.796 4188328.078
1406719.780
>> 4584890.778 4178864.584 1419611.655 4589354.898 4180122.057
1426529.969
>> 4586084.915
>> dst=. 10 3 $ 4190406.336 1419205.511 4578987.084 4198058.784
1411128.221
>> 4574636.297 4197986.816 1417726.961 4572625.286 4192437.202
1426034.821
>> 4575147.692 4185986.869 1425972.859 4580910.112 4184378.098
1410605.362
>> 4587284.395 4186367.950 1417240.794 4583258.087 4188266.426
1406788.485
>> 4584895.125 4178802.915 1419680.300 4589359.234 4180060.331
1426598.633
>> 4586089.223
>>
>>
>> --
>> Zoltan
>>
>>
>>
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>>
>
>
>
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