Hi,
Are there Carlson Survey users and professionals here?
I am new in Carlson Survey. Therefore I can't resolve my task.
There are 64 geodetic control points (benchmarks).
And I have got their coordinates in two systems: 1) in SK-42 (it is Pulkovo-42 Gauss-Kryuger Zone-9) and 2) in UTM-39 WGS-84.
My questions:
1) How can I find the 7 Helmert transformation parameters from first to second system in Carlson Survey?
and
2) After it, how can I transform (appling the parameters) my maps (drawings) from first system to second in Carlson Survey?
Thank you!
Binnat Khalilov
Did you ever find an answer to this?
No, Schneider. Nobody didn't answer to this.
Do you have any idea?
bingeomap, post: 337253, member: 10377 wrote: No, Schneider. Nobody didn't answer to this.
Do you have any idea?
No, I'm sorry I don't but, it was an interesting question.
I have find the functions for it....
But I don't know:
1) Are these 7 parameters for the global geocentric cartezian systems or local?
2) And can I use these parameters in other - thirty-party softwares or is it only for use in Carlson Survey?
bingeomap, post: 337255, member: 10377 wrote: I have find the functions for it....
But I don't know:
1) Are these 7 parameters for the global geocentric cartezian systems or local?
2) And can I use these parameters in other - thirty-party softwares or is it only for use in Carlson Survey?
From information I could find on the Carlson http://files.carlsonsw.com/mirror/manuals/Carlson_2016/source/General/Points/CooRDinate_File_Utilities/CooRDinate_File_Utilities.htm&apos ;">cfutransform routine:
[INDENT=1]The Helmert 7-Parameter method can also be used for local transformations. The 7-Parameter Values can be calculated from control points or entered by the user.[/INDENT]
[INDENT=1]
The Transformation Type chooses between doing a 2D transformation and 3D transformation. For the 3D transformation, the program transforms the x/y using the same method as the 2D transformation, and the z is transformed using an elevation difference model that is modeled by either a best-fit level plane or tilted plane as set by the Vertical Transformation setting.[/INDENT]
Since the process generates seven parameters of:
- Translate (X, Y, Z)
- Rotate (x, y, z)
- Scale
it would be an educated that these values can/could be used in other software applications that accept the Helmert 7-Parameter methodology. For what it's worth, I did find an http://www.engineeringsurveyor.com/software/transformations.xls&apos ;">Excel transformation spreadsheet at http://www.engineeringsurveyor.com&apos ;">www.engineeringsurveyor.com. I can't speak to its validity but it could be useful as an additional check. I hope this information helps.
Thanks, Ladd.
I understood, yet.
I'm afraid that a 7 perimeter transformation is more than I have ever understood. Yet, I suspect it has been done.
Without understanding.
N
Nate The Surveyor, post: 337401, member: 291 wrote: I'm afraid that a 7 perimeter transformation is more than I have ever understood. Yet, I suspect it has been done.
Without understanding.
N
Nate, what didn't you understand?
I did it, not knowing that was what I was doing. I knew it as "Best Fit" and did my own fuzzy math to make it do what I wanted. Thanks for posting that explanation.
Helmert transformations are based on the Euclidean grid system of the coordinates being used. A seven parameter system can be ECEF, but only if the source and destination coordinates are ECEF.
The basic work flow of Helmert transformations is:
1. Determine centroid of both systems. This is a simple average of each dimension, ie Nothing, Easting, Up. The translation will be from the average NEU of the source coordinates to the average NEU of the destination coordinates.
2. Determine radial inverse, direction and distance, from centroid to each point.
3. Determine scale factor by solving least squares of distance residuals of the radials.
4. Determine rotation by solving least squares of angular residuals of radials.
5. Apply best translation, rotation(s) and scale to source coordinates. Solve difference between transformed source and destination coordinates to see goodness of fit.
I should clarify No. 4. Actually it isn't a least squares of the angular residuals (in angular units), it is a least squares solution of the arc distance of the rotation angle x radial distance from centroid to point. This way points that are very close to the centroid and may have large residuals in angular units don't skew the results. By solving for the arc distance and linearizing the residuals, a scale factor can be determined, just like the scale factor for distances, that is then applied to the rotation angle (in angular units).
Shawn, thank you very much for detailed explanation. I realy didn't know these.