PART 1. Answers module 6: 'Transformations'

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1 Answers module 6: 'Transformations' PART 1 1 a A nominal measure scale refers to data that are in named categories. There is no order among these categories. That is, no category is better or more than another. b Pixels, i.e. picture elements. The coordinates refer to the column and row indices in the image / raster. The image has 918 columns and 567 rows. c - 2 a Control points or correspondence point are points that you have identified in both an input data set and a reference data set and for which you have acquired coordinates in both coordinate systems that are used for the estimation of the coefficients of the transformation function. Control points are also used to determine the positional accuracy of the georeferenced image. b You use control points to fit a polynomial function to the coordinate values of control points. Thus, the output coordinate values are an estimation of the true values and they are estimated by the polynomial transformation function! There will always exist a difference between estimated map coordinate based on the transformation function and the true map coordinate. This difference is called the residual. The formula 7.7 in Chang (page 122) is used to calculate the residual error. The residual is a measure for the goodness of the fit of the transformation function to the coordinates of the control points. A good fit results in low residuals. What happens is that the control points are used to determine the coefficients of the transformation functions. Then the map coordinates of the control points are estimated and compared with the true coordinates (the coordinates derived from the reference data set). The difference between true and estimated coordinate is the error. The x-error is calculated; the y-error is calculated. Both errors are squared and added. Then the square root is taken. The result is the residual of one control point. c No, two points are not enough to calculate the residuals. You need three points to get the coefficients of a 1 st order polynomial. With 4 points and more you can calculate residuals. d An acceptable Total RMS after the collection of 12 control points is between 2 and 4. f The RMS goes down; the fit of the polynomial function gets better. g A low RMS error (see above) should not be confused with an accurate rectification. A low RMS error does not automatically yield an accurate rectification. For example, using the minimum number of control points required by the mapping function, zero RMS error should be reported. Nevertheless, the transformation may contain substantial GRS Answers Module 6

2 errors due to a poorly entered control point. In fact, the RMS only indicates how well the transformation function could be calibrated to the control points. 3 a Nearest neighbor: the output cell gets the value of pixel that is nearest to the computed coordinates in the input raster; Bilinear interpolation: the output cell obtains values that are computed by bilinear interpolation (spatial averaging) over the several pixel values that are near to the computed coordinates in the input raster. b With nominal data you should use nearest neighbor resampling. Else you may be averaging class indices, e.g. the average of class 3 (grassland) and class 9 (buildings) is class 6 (water). This is nonsense of course. 4 a Near the centre of the image y coordinates are fine, but moving away from the centre there is a displacement in Y. The displacement appears to be curvilinear as a result of the distortion in the scanned map. 5 a The total RMS error should be lower. In the example used in Figure 7, the total RMS decreased from 2.78 to 1.08 meters. b Yes, when you zoom in you can observe a better overlap of the roads and field boundaries. c Yes, the image contained a curvilinear displacement. Linear (1 st order, Affine) transformations are not able to correct for these kind of distortions. A 2 nd order transformation function can correct curvilinear distortions, which should result in a lower RMS error. 6 a 1.9 meters, for example. The positional accuracy of the referenced image is about two meters. b The RMS error of the validation is likely larger than the RMS error of the transformation function. Mind that this answer refers to a validation whereas the previous referred to a calibration of the polynomial transformation function (how well the transformation function is fitted to the coordinate values of the control points). c For example: digitizing precision, zoom factor applied, choice of control points, resolution of input image, how well could control points be recognized/identified, order of the polynomial, type of distortion/displacement in the input image. GRS Answers Module 6

3 7 PART 2 When the points or polygons fall within different (raster) cells, which cell(s) get the value of the point or polygon data, or when 2 or more different point features situated in one raster cell? 8 a Ordinal, difference and order between the values. b DESCRIPTION VALUE WEIGHT good drinking water livestock drinking water poor drinking water toxic water 4 2 c In vector, geographic phenomena are defined by sets of coordinates. These coordinates are explicitly stored. For example a polygon is basically a set of coordinate pairs that define the boundaries of the enclosed area. A raster dataset is basically a matrix of cells. The location (coordinates) of a cell is determined by its location within the matrix. In other words, raster coordinates are contained in the ordering of the matrix, as opposed to the vector structure that stores coordinates that describe geographic phenomena. Thus, when you transform vector data to raster data the coordinates that describe the vector features disappear and are replaced by raster coordinates. GRS Answers Module 6

4 d 9 - cells with 'good' drinking water get value 1 (green) - cells with 'livestock' drinking water, but no good drinking water get value 2 (blue) - cells with poor drinking water, but no livestock and good drinking water get value 3 (orange) -cells with toxic drinking water, but no good, livestock or poor drinking water get value 4 (red) Cells with interstate highways had the highest weight, country roads the lowest. Cells with interstate highways became black (value 1) Cells with state roads, but no interstate roads, became dark grey (value 2) Cells with only country roads became light grey (value 3). GRS Answers Module 6

5 10 In this example the weight of the heather was the highest and the weight of the other land cover classes was equal. For raster cells with more than one polygon the most occurring polygon was sampled (except when it contained heather, than it sampled heather). You could discuss whether the lower right cell should be sampled, since it mostly contains no data /white. But then you should define a threshold value, which was not the case in this example. 11 a Because in the defined raster, the center of some of the raster cells didn t overlap with the polygons. B Value 2 has the highest number of cells (84), in total there are 244 cells ( ). C They are not empty cells. The value of these cells is NoData. D All cells that have the value of NoData are displayed in color. You can see that the raster dataset does not have the irregular outline of the vector dataset but is a rectangle. A substantial part of the raster is filled with value NoData. E The raster boundaries are defined by the most northern, eastern, southern and western point of the polygon dataset. F The total number of raster cells (incl. NoData cells) is 360. G Value 2 has the highest number of cells: The raster dataset contains in total cells (incl. NoData). H The one with the highest resolution (10 m cell size). This map gives the best approach of the outline of the vector dataset layer. Discrepancies between the 10 m raster and the original vector dataset layer are much smaller than the discrepancies between the 100 m raster and vector dataset. GRS Answers Module 6

6 12 a Is one of the solutions. All red/yellow features belong to different polygon features with value 1 [in this case 6 objects], all green/blue features belong to different polygon features with value 2 [in this case 4 object]. It assumes that an object is the same if they share a side. However another solution could be 2 objects with value 1 (red) and 1 object with value 2 (green). An object is created when a corner has the same value. Other options could also exist. Corners for objects with value 1 and sides for objects with value 2 and vice versa. The two solutions given here also use tight edges, where more curly edges could also be a possibility. A decision rule, stored in the software, decides which possibility will be taken. 13 a FID, Shape, ID, GRIDCODE, Soilcode b Soilvector1 has fewer fields, because they got lost in the transformation (question 11). During vector raster transformation all fields are lost except the field that contains the values that are used to determine the raster zones (values). Soilvector1 has one field that GRS Answers Module 6

7 Soil_types doesn t have (gridcode). c Along the feature boundaries. d Conversion process raster-vector: the process of converting a raster made up of cells into a dataset described by lines and polygons. 14 a Soilvector2 is more precise/accurate than Soilvector1, due to a smaller cellsize (10m, compared to 100m). The largest discrepancies are located along the Rhine and the north and east border. There are still discrepancies between Soilvector2 and the original dataset Soil_types. When you convert a vector dataset to a raster dataset some of the positional accuracy is lost. When you convert the raster dataset back to vector, the result will be less accurate (has a lower positional accuracy) than the original dataset. b The biggest discrepancies are located along the borders of the Rhine. 15 a There is no difference in geometric meaning between both rasters. Although the cell size is smaller the new raster is not more accurate since the source data was of a larger scale. b There is a difference in geometric meaning between both rasters, Although the cell size is smaller the new raster is not more accurate, since the source data was of a larger scale c No, if you resample categorical data you should not average them d If you decrease both rasters to a cell size of 10 meters you won t change the geometric meaning of the rasters. You should however keep in mind that you are comparing rasters with a cell size of 30 and 50 meters not rasters with a accuracy of 10 meters An alternative would be to resample the 30 meter raster to a 50 meter raster. The data is again categorical, so you can only use nearest neighbor or minority, majority methods. In the latter case you won t find differences that are created due to more accurate mapping (e.g. roads which were not presented on the map of 1950). GRS Answers Module 6