Transform Job to Control

This option will use a set of Control points that you enter the (AMG) co-ordinates.
Create Control points in the job usinge Points/Control menu option.

Overview

The program will use the new (AMG) co-ords and the existing co-ords to create a helmert transform matrix. This transformatiuon will be used to shift/rotate/scale the job onto the final co-ords.
Transform Job to Control Option. This option will find all the Control points in the current job and will display the list of points for the user to select which control points to use for the transformation. You can enable or disable control points in the list which will be used to compute the transformation. You need a minimum of two points, and can have a maximum of 100 points.



1. Control Points. Use the Points/Control dialog to create the control points.

2. Use the Check Control Fits the Job button to search the control file for the control co-ordinates for those points. It will use a “Least Squares” procedure to compute parameters to rotate, scale and shift the co-ordinates and will apply these to the job control points.

The difference between the transformed values and the Control values will be displayed in the window. This enables the user to check how the control fits and isolates any erroneous control values.

Points can be made active and inactive as appropriate and the transformation recalculated until you are satisfied that the control is satisfactory.

3. Transform Job to Control If the control differences are small enough you can use the Transform Job to Control button to use the computed transformation parameters to transform every point in the job onto the new co-ordinates system.

Enter a listing file if required and the transformation scale, rotation and control point residuals will be written to the listing file when you do a ‘Check Control Fits the Job’.

Use Close to exit from the transformation screen and if the job is then on a geodetic datum, use the Change Zone option to set the new datum and zone.

Automatic Rejection of Bad Control Points

The transformation may not use all the control points which you select.

If more than four points are enabled, the program will check the control point with the largest residuals and if that control point is more than 2.5 standard deviations from the rest, it will automatically be excluded. Points more than 2.5 SD from the others have about a 1.5% probability of being part of the set.

How It Works

The transform uses the control values to compute the parameters for a helmert transformation between the two systems. This provides for a shift in X , a shift in Y a rotation and a scale change.

If two control values are used you get a unique solution. If more than two control values are used there is redundent data and a least squares procedure is used to calculate the parameters.
It is recommended that more than two control values are used so that you can get a measure of the way that the control values fit with the job data.

After the transformation matrix is computed, the job coordinates for the control points are transformed and compared to the actual control values. The differences are shown as residuals in the report.

Point   Point  Point           Final        Final    ---------Residuals--------
Active     No  Name          Easting        Northing       dX        dY        dZ
 Yes     1202  3             458489.888   8145233.016     0.423     0.362   -19.991
 Yes     1203  148           455342.389   8145170.582    -0.152     0.087        
 Yes     1492  2             458496.164   8146285.149    -0.271    -0.448   -19.660

As you can see, point 3 has residuals of 0.423 in X and 0.362 in north and point 2 has residuals of -0.271 in X and -0.448 in Y So there is a missmatch of a couple of metres between the shape shown by the control and the shape for the same points in the survey. If you look at the actual distances between the points in both systems you will see that the distance 1203 - 1492 only about 100mm different from the control distance between these points, but 1202-1203 and 1492- 1202 each have differences of over a metre ane each is in adifferent direction. So, it appears that point 1202 may be a bit suspect.

However, if you have no other information and have to adopt all values, each point in the job will be transformed by the transformation matrix. The control point however will simply adopt the control coordinates and not the transformed values. Because of this a point which started say 1mm from point 3 would be about 0.27 in X and 0.45 in Y from that point after transformation.