1. Problems with Vector Overlay Analysis (esp. Polygon)
Overlay analysis using the vector spatial data model is highly computationally intensive:
Complicated input layers can tax even current processors
There is a tradeoff between the complexity vs. the interpretability of results
Complex input layers with many polygons can result in many more polygon combinations… can we make sense of all those combinations?

2. Problems with Vector Overlay Analysis (esp. Polygon)
There are often spatial mismatches between input layers
Overlay can result in spurious sliver polygons
We can “filter” out spurious slivers by querying to select all polygons with AREA less than some minimum threshold
It is difficult to choose a threshold to avoid deleting ‘real’ polygons

3. Boolean Operations
Boolean operations of OR & AND correspond to UNION & INTERSECTION, used in vector-based analyses A B A B OR A B
UNION A B
AND A B
INTERSECTION
We can apply these concepts in the raster spatial data model, when two input layers contain true/false or 1/0 data:

4. Boolean Operations with Raster Layers
The AND operation requires that the value of cells in both input layers be equal to 1 for the output to have a value of 1: 1 1
AND =
The OR operation requires that the value of a cells in either input layer be equal to 1 for the output to have a value of 1: 1 1 OR =

5. Algebraic Operations w/ Raster Layers
We can extend this concept from Boolean logic to algebra
Map algebra:
Each cell is a number
Mathematical operations are applied to the layers, calculated on a cell-by-cell basis
The result for each cell is placed in a new layer.
Example: suitability analysis
Multiple input layers determine suitable sites:
The pixels attributes in each layer represent ‘scores’.
Layers are weighted based on their importance.
Layer scores are added on a per-pixel basis.

6. Simple Arithmetic Operations1 + = 2
Summation 1  =
Multiplication 1 + = 3 2
Summation of more than two layers
Near the mall
Near work
Near friend’s house
Good place to live?

7. Raster (Image) Difference5 1 7 6 3 4 2 - = -2 -1 -3
The difference between two layers:
An application of taking the differences between layers is a form of change detection:
Example: you have 2 images of the same forest, 10 years apart. To measure forest growth or cuts, you can take the difference in infrared light between image dates.
More infrared light = more chlorophyll and more greenery
Question: How can the locations where a substantial changes have occurred be identified using the difference layer?

8. Raster (Image) Division
Question: Can we perform the following operation? Are there any circumstances where we cannot perform this operation? Why or why not?  =

9. More Complex Operations
Linear Transformation 2 3 5 1 4 + = a b c
This could applied in the context of computing a statistical linear regression (e.g. output y = a*x1 + b*x2 + c*x3) using raster layers
Example: suitability analysis.
Suitable place to live example.
Near the mall – low importance (a=1)
Near work – high importance (a=4)
Near friend’s house – moderate importance (a=3)

10. Problems with point location data
You have point location data, but you want data for your whole site/region.
Examples:
Air temperature maps created from point data.
Water pollution levels measured at points.
Solution:
Estimate areas with no data.

11. Spatial Interpolation
You have point data (temp or air pollution levels).
You want the values across your full study site.
Spatial interpolation estimates values in areas with no data.
creates a contour map by drawing isolines between the data points, or
creates a raster digital elevation model which has a value for every cell

12. Spatial Interpolation: Inverse Distance Weighting (IDW)
One method of interpolation is inverse distance weighting:
The unknown value at a point is estimated by taking a weighted average of known values
Those known points closer to the unknown point have higher weights.
Those known points farther from the unknown point have lower weights.

13. Sample weighting function
Spatial Interpolation: Inverse Distance Weighting (IDW)
point i
known value zi
distance di
weight wi
unknown value (to be interpolated) at
location x
The estimate of the unknown value is a weighted average
Sample weighting function

14. Issues with IDW
Weighted average estimates are always between the min and max known values.
If the known (sampled) points did not include the minima and maxima (e.g., mountain peaks and valleys), your data will be less extreme than reality
It is thus important to position sample points to include the extremes whenever possible

15. Issues with IDW The dashed line is a hill.
The x’s are sampled elevation points.
The black line is the interpolated (estimated) hill elevation. With IDW, the unknown points tend towards the overall mean.

16. Triangle Irregular Networks (TIN)
Most often used for elevation surfaces.
Points are known data values (of elevation).
Lines are drawn between nearby points to create a set of irregular triangles.
The value of the variable (e.g. elevation) moves along the lines evenly from one point to the next.
The triangle between 3 lines represents a flat surface with slope and aspect.
With the TIN, the feature’s value every point on the TIN is estimated.