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The resulting areas are the ones that an insurance company may want to target to sell flood insurance. City planners could also use this information for disaster planning services.
(2) Examine soil samples of a farm area using interpolation
In this step, you will help a farmer balance the pH levels in a field that is being prepared for the next growing season. Some crops have better yields when the soils have a balanced pH level. Seven is a balanced pH level. Areas in the field with a pH less than seven are treated with lime (limestone) to raise the pH level. Areas with a pH above seven are treated with ammonium sulfate to lower the pH level.
The farmer should not treat an entire field with all lime or all ammonium sulfate, but should locate areas of high and low pH and treat them accordingly (precision
farming techniques). You will help the farmer find the areas that should be treated with ammonium sulfate (areas with pH greater than seven).
Look at data - soil samples in farm area
This map contains two layers. The Soil samples layer represents the soil samples that were collected in the field and tested for chemical composition. It contains several fields containing the chemical levels at each sample point. The Farm field layer represents the extent of the farmer's field.
For this analysis you will follow the steps shown in the flow chart below.
Isolate areas of high pH. Next, you will isolate the high pH areas by creating a layer containing only areas with pH levels above seven.
Raster Calculator - [pH surface] > 7, pH Treatment Areas. Values of 1 are those areas that where pH is greater than seven.
values of 1 already isolated
The pH treatment areas are the areas that the farmer should treat with ammonium sulfate to lower the pH to seven so that it is balanced. The farm size is about 5.35 acres (233,046 square feet or 21,650 square meters) and the combined size of the newly defined treatment areas is about 0.145 acres (6, square feet or 588 square meters).
If the ammonium sulfate treatment costs $50.00 per acre, treating the entire 5. acres costs about $267.50, while treating 0.145 acres costs about $7.25. Treating only the areas that actually need it results in a possible savings of $260.25. Imagine if the farmer had several fields.
Farmers may use similar techniques when applying fertilizers and pesticides to their fields. Also, histories of crop yield and treatment can be mapped over time and used for future planning.
The map contains three layers: Shops, Customers, and Streets. The Shops layer contains the locations of existing coffee shops. The Customers layer is not turned on; you will turn it on later.
Examine the locations of the existing shops. For this analysis, you will assume that any shops within 1 mile of each other will compete for customers. Potential sites for a new shop should therefore be more than 1 mile from any existing shops.
For this analysis you will follow the steps shown in the flow chart below.
You will start the analysis by creating a surface representing the distance from the shops. You will then create a surface representing the density of customer spending. Finally, you will query the distance and density layers to find the areas that are a mile or more from existing shops and with high spending density.
Use straight line distance function - dialog
A distance surface is created and added to the map. Areas shown in yellow and orange are close to the shops, while areas shown in purple and blue are farther from the shops.
For Input data, make sure that Customers is selected. For Population field, click SPENDING. Click OK. A spending density surface is created and added to the map.
The dark blue areas have the greatest spending density of customers.
Want to find areas that are more than 1 mile (5,280 feet) from an existing shop and that are in a high spending density customer area.
Raster Calculator. ([Distance to Shops] > 5280) & ([Spending density] > .02).
