> U@ @bjbj 7\\\8$<Q444LPPPPPPP$SRUQQBBBQ4L7VQ!!!B84LP!BP!!6"LXP4A)<\NN(PlQ0Q=NhV$!vVPPVP,'r!\MQQd\!\[Slide Title..]
Introduction to Geographic Information Systems Distance Learning Version
Lecture 4 Script
Introduction: Hello and welcome to Lecture four of the distance learning version of Introduction to Geographic Information Systems. By this time, you should be so excited about coming to our regularly scheduled sessions that you can scarcely contain yourselves. Im going to work hard to channel this exuberance and excitement. There are some advantages to living in an enchanted virtual world. In any event, this lecture will be important because some of the data analysis operations discussed here, you will use in completing your semester project. (Next Slide Please)
[Slide Intro...]
This lecture will deal with turning data into information. Up to this point weve been obsessed with the data, both spatial and attribute. This is the first time we are going to talk about actually performing operations on the data to solve problems and answer what if questions. To do that, we are going to have to get into some detail and actually use some mathematical expressions.
The first item on the agenda is how to measure distances in the vector and raster worlds. (Next Slide Please)
Then, well look at attribute queries, which weve seen before, but now well see how to form multiple questions in one query. (Next Slide Please)
Next, well see how to answer the question, How close is it? by performing a proximity analysis. (Next Slide Please)
If we actually overlay one layer or map over another mathematically, we can answer questions and produce new maps that contain the characteristics of the original maps. For instance, from our Ski Resort example, if we overlay the layer containing the meteorological stations with a land use map of forest and non-forest areas, we produce a map which tells us which meteorological stations are located in the forests, and non- forest areas. (Next Slide Please)
And finally, the analysis of surfaces, like digital terrain models, or DTMS, and networks will be discussed. As usual, we start out with definitions. (Next Slide Please)
[Slide Database Terminology ]
Take a moment to look at the table. Weve seen the top six before, but it doesnt hurt to review them. (Pause) Lets pay particular attention to the terms in the last two rows. A function or operation is a data analysis procedure performed by a GIS but most importantly, formulated by YOU. Again, let me remind you that you will need to use data analysis techniques to complete your semester project.
And lastly, an algorithm is a series of steps to solve a problem. Its a road map or flow chart that shows how the problem will be solved. Here again, this has important implications related to your project. Lets look at several GIS operations, starting with the most fundamental one measurements. (Next Slide Please)
[Slide Measurements...]
We need to be able to measure lengths of lines, and perimeters and areas of regions. The methods we use will be dependent on whether the layer is a vector or raster layer. (Next Slide Please)
[Slide Vector GIS Measurements...]
Here we see that in vector space, we use the Pythagorean Theorem to solve for the length of lines. The perimeter of an area is determined by adding the length of the lines forming the polygon for the area. Remember when a topology table was added to vector data? Now, to calculate the perimeter of a polygon, we just need to add up the lengths of the lines that make up the perimeter of the polygon. To calculate areas, triangles are either added or subtracted to form and calculate the area. Im sure youve seen this stuff before in math or physics. (Next Slide Please)
[Slide Raster GIS Measurements ]
In raster space, the unit of measure is the length of the side of a cell. In the case of LandSAT Satellite images, the length of a cell is 30m. Then, calculating the length of a straight line from the corner of one cell to another is given by the Pythagorean Theorem with the distances calculated by the number of cells times the dimension of a cell. There is another way to define distance what is called a Manhattan distance - so named because of the way you have to go from point A to B in a city by walking along block boundaries. Theres still another way to measure distance called proximity distance. In this scheme, concentric circles radiating from point A are produced in this case at equal intervals of the units of the cell. Any point like point B can be interpolated exactly from the contours of equal distance.
As for perimeter and area for perimeter we just count up the number of sides in the perimeter and multiply by the length of a cell side. For area, we count the number of cells and then multiply by the area of a cell, which is the length of the side squared. Thats about it for measurements pretty straight forward. (Next Slide Please)
[Slide Queries....]
Weve seen queries before. Remember the SQL language? - The Standard Query Language? We saw that we could use the SQL to look at the database and retrieve data. Queries can also answer the questions How many? and Where are they? Well add the power of SQL by looking at the notion of writing several queries in one expression. To do that, we need to look at Boolean Operators. The idea of using Boolean Operators is most applicable to vector models since the database is organized with many attributes stored in tables, unlike a raster model. (Next Slide Please)
[Slide Boolean Operators...]
To study Boolean Operators, lets return to our Ski Resort example. Lets let A be the set of Luxury Hotels and B be the set of all hotels with more than 20 rooms. Using Boolean Operators, four questions can be posed. (Next Slide Please)
[Slide Boolean Operators Continued...]
Here are the four questions that can be asked and answered using Boolean Operators, which appear in red font. Lets read these then diagram them to see them better.
Which Hotels are Luxury and have more than 20 bedrooms?
Which Hotels are Luxury or have more that 20 bedrooms?
Which Hotels are Luxury but do not have 20 or more bedrooms?
Which Hotels are either Luxury or have more that 20 bedrooms, but not both?
So we see that there are four Boolean Operators: And, or, not and xnot.
Next Slide Please)
[Slide Boolean Operators Venn Diagrams ...]
These are the Venn diagrams from the four statements on the previous slide.
1. Which Hotels are Luxury and have more than 20 bedrooms?
Looking at the AND operator, the cross hatched area represents that sub set that satisfies both criteria. The expression written in SQL is Hotel=Luxury AND Bedrooms >20.
Which Hotels are Luxury or have more that 20 bedrooms?
If we look at the OR operator, then all Hotels meet the criteria.
Which Hotels are Luxury but do not have 20 or more bedrooms?
For the NOT operator, the cross hatched area represents all Hotels that are luxury but do not have more than 20 bedrooms.
Which Hotels are either Luxury or have more that 20 bedrooms, but not both?
The XOR operator is the inverse of the AND operator. The cross hatched area represents Hotels that are either luxury with less than 20 bedrooms, or other hotels with more than 20 bedrooms. At first glance this stuff looks complicated, but its really very logical and helpful in forming complicated queries. But, we have to be careful using Boolean Operators, because we can get drastically different results if we use an AND operator instead of an OR. Queries in Raster space can take on different forms. (Next Slide Please)
[Slide Queries...]
We can perform a query in a raster domain using a process called reclassification. Reclassification can produce a Boolean image (basically ones and zeros or black and white). Lets look at an example in a land use image, one of the classes of land use is forest. If we wish to produce a layer with forest only areas, we can reclassify the land use image. Keep in mind; we could do this by writing a query. (Next Slide Please)
[Slide Bloomfield Land Use..]
This is a raster land use map of Bloomfield, CT. Note the large patches of green, which are denoting the forests. We can reclassify all the cell values that represent the forest cells to one, and then reclassify all the non-forest cells to zero. When this is done, we obtain a Boolean map or a map that has two attributes - ones or zeros. (Next Slide Please)
[Slide Bloomfield Land Use Only Forest...]
This is the Boolean image formed by the reclassification. We can easily see now where the forested areas are. Its clear that reclassification produces the same final product as writing a query except that it may have taken several steps to produce this Boolean image, where reclassification accomplished this in one step. It is usual practice to show a Boolean image in black and white. (Next Slide Please)
[Slide Proximity Analysis ...]
Now we get to ask the question How close are things? The most common procedure that is used to answer this question is Buffering. Buffering is a GIS operation that creates a zone of interest around an entity or set of entities.
(Next Slide Please)
[Slide Buffer Zones ]
Buffering is an easy process to visualize, but a lot more difficult for the software to execute. We can see possible buffer zones around points, lines and areas. Buffers can be simple areas around a point or line, or a common border around an area seen on the left, to a more complicated buffer zone as seen on figures on the right. Buffers by themselves can be interesting, but their real value involves combining them with other layers in a process called overlaying which well talk about in a few minutes. As an example of the use of buffering, suppose that we are studying the location of radioactive waste disposal sites. Two of the criteria are that the sites must be located more than three km from a railway, and be situated on clay soil because clay is resistant to liquid penetration and can be used as a natural penetration barrier, in case of a leak. The first step is to create a buffer zone around the railways, which is easily done in a GIS. (Next Slide Please)
[Slide Buffer Zones 3 km...]
Here the buffer zones are drawn in blue around the rail lines in red. Note the blending of the buffers in areas of overlap. This blending is also supplied by the GIS software. Well see these buffer zones used later in the waste disposal example. In raster images, buffers are created by the proximity process, which was presented in the measurement section. Recall that proximity distances radiate from a point and produce contours of constant distance from the point in question. If there are several points to buffer, the software will again blend the proximity zones where they overlap.(Next Slide Please)
[Slide Proximity Map for Hotels in Ski Resort. ]
If we want to see how close the individual Hotels are in the Ski Resort, we can perform a proximity analysis and create a surface called a distance surface. In this image, the darker the color, the closer the cell is to the Hotel, which is represented by a white cell. Note how the software blended the image as the distance surfaces for each Hotel overlap. If we want to see a particular buffer zone of say 125 km, then we can reclassify the distance surface image making all distance less than 125 km = 1 and given the white color, and all other cells assigned black in a conventional Boolean image. (Next Slide Please)
[Slide Overlay Operations ]
Perhaps one of the most import data analysis techniques is called overlaying. In GIS, overlaying has two meanings. The first youve already seen in ArcView when you drew one layer on top of another. In this lecture, we will take the second definition, which is of course related to data analysis that is, a GIS operation that combines information from two layers into a new layer. Again, well look at vector and raster overlay operations separately.
(Next Slide Please)
[Slide Vector Overly Operations..]
An important condition for the overlay process to work is that all layers must be topologically correct. In other words all polygons have to close, all lines must meet at points, and so forth.
When the lines and polygons from the original layers are overlayed, new lines and polygons are formed. The software has to then assign attributes to the new entities, after it actually forms the new entitles. This process is governed by the laws of geometry, and takes some computational power. There are three types of overlay operations that are possible in a vector layer. (Next Slide Please)
[Slide Vector Overlay Types]
The three types are point in polygon, line in polygon and polygon in polygon. Well look at each one in detail, again using our Ski Resort example. (Next Slide Please)
[Slide Point in Polygon ..]
If we want to find out which meteorological stations are located in the forest, and which are not, we can overlay the point layer of the Meteorological stations on the polygon forest layer. This produces a new point layer in which the points now have new attributes that describe whether or not they are in the forest. The line in polygon case is a little more complicated. (Next Slide Please)
[Slide Line in Polygon..]
Now, if we want to know which roads, or parts of roads are in the forest, we can overlay a line layer on a polygon forest layer. The result is a new line layer with the line segment labeled with whether or not they are in the forest. Note the complication here the new layer has new line segments created by the overlay process. The original line 1 now has two segments. The new numbered segments 1 and 2 are part of the original line with 1 in the forest and 2 in the non-forest. When we go to polygon in polygon things get interesting. (Next Slide Please)
[Slide Polygon in Polygon..]
The polygon in polygon process resembles the Boolean operator method used in the query procedure.
Staying with our resort example, if we want to know which new areas are within the forest/non forest map or within the resort map, we would use the OR Boolean operation. In this case all of the new polygons fall into these categories. In GIS operations, this is called a UNION.
Next we look at the question Where are the areas of forest and non forest inside the resort boundary. In Boolean, this would be resort NOT forest/non forest. In GIS talk, this is an IDENTITY process. The question it answers is Where in the resort, are the forest and non forest areas?
And finally, we would use the AND Boolean operator to answer the question Within the resort boundary, where is the forested area? In GIS language, this is called an INTERSECT process. (Next Slide Please)
[Slide Vector Overlay Rail Buffer Zone and Clay Geology..]
Remember the buffer zone example for the radioactive waste problem? We saw the buffer zones drawn around the rail lines. Now if we overlay the rail buffer zone with the clay geology layer using an INTERSECT process, we get the clay areas in yellow that are within the rail line buffers. Any site within the yellow areas is not acceptable. Conversely, clay areas not in yellow are acceptable. (Next Slide Please)
[Slide Little Grey Cells Quiz]
OK Heres our Little Gray Cells Quiz. I may ask some of you for your answers in the chat session, so you might want to jot down some notes. Ill give you a few moments now to come up with some answers. (Pause here)
Do Not Read
A raster image is made up of cells. T or F
Which Boolean operator will allow both input layers to exist simultaneously?
The creation of a zone of interest around an entity, or set of entities is called an overlay.
T or F
(Next Slide Please)
[Slide Break]
OK - Lets take a short break to stretch, wake up, or whatever.
(30 to 60 second break) (Next Slide Please)
[Slide Raster Overlay Operations]
In many ways, overlay operations in raster space are much easier to understand than their vector counterparts. We have to keep a couple of things in mind: in raster images, points, lines and areas are all represented by groups of cells; it is also important to understand the kind of data in the cells, such as - is the data format ratio, interval, and so forth.?
To overlay raster layers, we can use the math operators plus, minus, multiply and divide. This process has a name map algebra. To study go-no go problems, sometimes Boolean images are used. Lets go through the raster equivalent of the vector overlay operations such as point-in-polygon. Well see that it doesnt make much difference in rasters if we overlay points, lines or polygons in polygon because the points, lines and polygons are just groups of cells. (Next Slide Please)
[Slide Raster Point-In-Polygon]
If we overlay the points of the meteorological stations (or met stations for short) on the resort area in an addition procedure, the result will be an image for which the cell values have been simply added together. In the result image we can interpret the outcome as a zero being neither a met station or in the resort; a 1 being a met station only, a 10 being a resort only cell, and an 11 having both a met station and in the resort. (Next Slide Please)
[Slide Raster Line-In-Polygon]
Looking at the line in polygon problem, we see a roads line layer overlaid on a forest area layer. Using the map algebra add process, we see that the results are very similar to the point-in-polygon problem. A seven cell value in the result layer would represent a portion of the road located in the forests. (Next Slide Please)
[Slide Raster Polygon-In-Polygon]
And finally we see the addition of a forestry area layer with a resort area layer or polygon-on-polygon. Again, using map algebra, we see that a 15 represents a forest and resort together. If we wanted just a map of just the areas where the forest and resort exist, we could reclassify the results map, setting every number but 15 to zero. There is another way to accomplish this result. (Next Slide Please)
[Slide Raster Line-In-Polygon +/x]
If we reclassify both images to Boolean form, then add them together, we get a result similar to the previous slide. In this case a 2 represents the areas that are both forest and resort. However, if we multiply the two images, we get another Boolean result image for which the forest and resort are 1 and all other combinations are zero. In this case its easy to spot the forest resort combination. In the previous slide, we had to determine that a 15 was the magic number. OK - I think you get the idea of how the overlay procedure works. Youve probably already seen the process in the lab exercises. The last topic to be covered involves the analysis of surfaces and networks. But before we look at how to analyze surfaces, the topic of spatial interpolation needs to be introduced, because it is interpolation that can actually form the surface we want to analyze. (Next Slide Please)
[Slide Spatial Interpolation ]
Spatial interpolation allows us to estimate, and I emphasize the word estimate, values at points on a surface where we dont have data or samples. In other words, interpolation allows us to fill in the gaps in the data. The most commonly used surface in Civil Engineering is the contour surface or DTM. In order to draw smooth contours of equal elevation, we will need to interpolate the raw data from the field. Most GIS software has interpolation routines available. There are several techniques for interpolation of data well look at three prominent ones. . (Next Slide Please)
[Slide Spatial Interpolation Techniques Thiessen polygons..]
The idea behind Thiessen Polygons is straightforward. Envision drawing a line between a data observation point and one if its neighbors. As a potential side of a polygon around the data point, draw a perpendicular bisector to the line, here shown dotted in red. Repeat the process to all neighboring points, and construct a polygon from the intersection of the perpendicular bisectors. The value of the observation point is then assigned to the polygon. If this process is carried out on all data points, a surface of polygons is formed. . (Next Slide Please)
[Slide Interpolated Elevation Thiessen polygons..]
The left hand image is the original elevation surface or the benchmark surface for this study. If you look closely, you can see the irregularly spaced sample points. The legend shows that the lowest elevation is black, and the highest is dark green. Using the Thiessen method, polygons are constructed around the sample data points and a polygon surface is formed as shown on the right hand image. We can conclude from observation that Thiessen polygon do not work well when trying to capture a continuous surface like a DTM. (Next Slide Please)
[Slide Spatial Interpolation Techniques Triangular Irregular Networks - TINS..]
We have already talked about the formation of a Triangular Irregular Network or TIN. Recall that a TIN is formed by triangular planes in space. Since we know the x, y, z coordinates of the corners of the triangle, we can write the equation of that triangular plane. Knowing the equation of the plane, for any horizontal position inside the triangle, we can easily calculate the elevation. If a line is drawn through points of equal elevation, a contour surface can be formed. (Next Slide Please)
[Slide Interpolated Elevation TIN..]
As just mentioned, by connecting lines of equal elevation as interpolated from the TIN, we can form an elevation surface. Note here that the TIN method produces a continuously varying surface, as is proper for a terrain model. We note that this image is much closer to the original elevation surface than the Thiessen polygon surface. (Next Slide Please)
[Slide Spatial Interpolation Techniques Distance Weighted Average...]
The last interpolation technique well look at is called the Distance Weighted Average method. In this method, we calculate the value at a given point and give more weight to the data points that are closest to it. As the equation shows, the value of - lets say an elevation at a point identified by the subscript zero - is calculated by summing in the numerator, the product of the elevation at a given data point times the square of the inverse of the distance of that data point from point zero. This is then divided by the denominator, which is the sum of the squares of the inverses of the distances. You can see that this model gives much more weight or importance to the data points closest to point zero, and greatly discounts the points that are furthest away. The one drawback to this method is that if you put point zero over one of the data points, you will not calculate the known value for the data point. So its not well suited for forming surfaces where the data is accurate, such as DTMs. It is better applied to situations where the data itself is known to have some error or uncertainty in it such as census data, or questionnaire responses. (Next Slide Please)
[Slide Interpolated Elevation Distance Weighted Average...]
Here we see the surface generated by the distance weighted average method. It looks somewhat different than the original surface from which the data points ere supplied. If we were to perform an error analysis, that is, compare each image we have generated with the original elevation surface, we would find that the TIN method produced the surface with the least errors. This is not surprising since TINS were set up to form the vector versions of DTMs. There are many more methods of interpolation, but we really cant afford to spend more time on this topic. Well, now that we can actually form surfaces using the techniques of interpolation, we can now look at performing analyses on these surfaces. (Next Slide Please)
[Slide Analysis of Surfaces]
Since a lot of Civil Engineering applications of GIS involve DTMs, we will look at two fundamental analyses that can be applied to DTMs: the generation and display of slope and aspect; and the use of visibility analyses to form viewsheds. Lets look at slope and aspect first. As usual, there are different methods for calculation of slope and aspect when applied to raster DTMs or vector TINS. (Next Slide Please)
[Slide Analysis of Surfaces Slope/Aspect.]
As always, we have to start with the definitions of slope and aspect. We can see a tilted plane in space outlined in red. Somewhere in that plane, there will be a line that has the maximum slope. If we take that line, S, and break it down into its horizontal and vertical components as shown in the top triangle, we can define the slope several ways: as the angle in degrees or radians; as the tangent of , rise over run or c/b; or in percent as 100 times the tangent of . Now, if we look at the projection of S onto a horizontal plane, the angle between the projection and a North Arrow is called the Aspect of the surface. We know that slope measures the steepness of the surface, and now we see that aspect measures the direction that the slope is facing. In a vector TIN, the calculation of slope and aspect is relatively easy because we know the equation of the triangular surface in space. For raster DTMs we need to do some preliminary work. (Next Slide Please)
[Slide Analysis of Surfaces Slope/Aspect.]
For a raster DTM, if we want to calculate the slope at a point on the surface, we place a 3 cell by 3 cell filter over the point, and use the elevation data in the cells of the filter to form a best fit plane, whose equation will be z = a + bx + vy. From this equation we can calculate the slope , and the aspect angle. (Next Slide Please)
[Slide Analysis of Surfaces Slope/Aspect& ]
We see the method just described applied to a raster DTM with 10m contour intervals shown. If we calculate the slope and aspect at each cell, two more surfaces are formed as shown. In the top image, a red line separates the south facing slopes from the north facing slopes. The demarcation is quite obvious for this surface. The slope image shows areas of high slope in light gray or white, and the flatter surfaces in the darker grays. Why are slope and aspect important? Slope calculations are important in determining surface water flow. We can imagine the flow of surface water down a slope until it intersects another slope. The intersection of the slopes forms a watercourse or stream, and the water continues down the course. By looking at the slopes in the area of the course, we can determine the watershed area that feeds the stream. This area is important in hydrological studies and youll hear more about this in Water Resources Engineering. As for aspect, this concept can be used to locate such things as solar panels, which must face in a certain direction, and using our ski resort example, we can determine which slopes around the ski trails are north facing, because north facing slopes will have a buildup of snow and pose a potential avalanche problem. There is one more analysis technique to look at that uses DTM data visibility analysis. (Next Slide Please)
[Slide Visibility Analysis]
Visibility analysis is simple in concept: if an observer stands on a certain spot, what can he or she see from that location, considering the surrounding terrain? If we draw a cross section of the terrain, and draw a line from the observer to some distant point, we can determine which areas in the terrain are above or below the line. Anything below an obstruction cannot be seen. This process is called ray tracing. If we repeat the ray tracing procedure 360 degrees around the observation point, we can form what is called a viewshed surface. (Next Slide Please)
[Slide Ray Tracing for Visibility Analysis]
Here we see a diagram of the process just described. Note the rays shown hit the peaks between the observer and the end point B, and the invisible or dead ground can be located. If the cells in the ray are filled in Boolean fashion, white for non-visible and dark for visible, an inter visibility matrix is formed. If we repeat this for other rays around the observer, then a Boolean surface is formed around the observer. (Next Slide Please)
[Slide Viewshed Analysis]
This is a viewshed analysis drawn over a DTM, which show the higher elevations in white and the lower evaluations in dark gray. The observer position is shown by the green dot near the bottom of the map. The viewshed Boolean analysis shows all visible areas in red and everything else is not visible. This analysis applied to a vector surface is similar in application. Some GIS packages allow the input of barriers such as tall buildings, and can even compute the extent of shadows for a given inclination angle of the sun. One good application of visibility analysis is the location of cell phone towers. Using this technique, dead zones can be identified. Ok - the last topic to be covered today is Network Analysis. As usual, well only touch the surface, no pun intended. (Next Slide Please)
[Slide Network Analysis]
Lets review what a network is a set of interconnected lines through which resources can flow. So rivers, roads, pipelines, and cables can all be identified as networks. For Civil Engineers, most network analyses will focus on road systems and Transportation Engineering in general. One of the most important parameters on a network analysis is impedance, which is defined as resistance to flow. In a road network, impedances can be the length of a link, turns, travel down one-way streets, or traveling over or under other links. One fundamental network analysis problem is to find the shortest travel path between two points. (Next Slide Please).
[Slide Network Example]
Here we see a symbolic drawing of a network of roads connecting six cities. The red numbers in parentheses are the travel times or impedances assigned to each link. Again, this is a simple example one in which you could find the shortest travel time between cities 1 and 6 by observation. But imagine that this network is a cross-country network of a thousand routes. We can see that a GIS network analysis program will be needed to look at all the possible combinations and permutations. (Next Slide Please)
[Slide Shortest Path Example]
We can set up an impedance matrix, with the city numbers as the rows and columns, and the data in the table representing the travel times between the individual cities. A computer algorithm could then start in the 1.1 position, and find all the points that connect to one, then go from these points to the next points, until a path is formed from 1 to 6. Here, we show two possible paths: 1 to 2 to 3 to 4 to 5 to 6. Adding up the impedances, it will take 110 minutes. Another possibility is 1 to 3 to 6. This path takes 72 minutes, which is the shortest path in time between 1 & 6. Things get a little more complicated if we try to account for the impedances assigned, say, to traffic lights. But, GIS software is set up to handle these complications. OK the last thing we are going to cover in this session is perhaps the most important, because it has a direct bearing on your semester project. Now that we have learned about data management and data analysis, how do we go about actually planning an analysis? The answer is to begin with a Cartographic model. (Next Slide Please)
[Slide Cartographic Model Formulation ]
We have looked at parts of this problem before, but now we can put the whole process of finding an answer together. The form that will describe the process is the cartographic model, which is a flow chart of the processes that we are going to have to use. The problem to be solved is find a site to store nuclear waste. The criteria for the sites are as follows:
Located on suitable soil or otherwise known as suitable geology
Has to be away from large population centers
Has to be away from major roads
Cannot be located in a park or conservation area
We note that at this point, its not really necessary to have all the details specified, such as: how far should the site be from the major roads?. (Next Slide Please)
[Slide Cartographic Model ]
Here is the cartographic model to solve the problem just described. We will work from the original data to the final map, which is the answer to the problem.
Beginning with the geology and population criteria, we first must pull the high population areas out of the population database. We can do this by reclassifying the data or by writing a query. Then, Overlay A is formed using suitable geology not in high population. Now, going to the roads data, we reselect out the major roads or write a query to select the major roads from all the roads. These major roads can be buffered to form the proximity to major roads layer. Then, Overlay B can be formed from Overlay A and the proximity to roads layers. Overlay B will contain sites that are in suitable geology, not in high population, and not in proximity to major roads. Now finally, if we form Overlay C by overlaying Overlay B and the conservation layer, we produce a final map, which meets all the original criteria. Note that there is not one correct cartographic model. We could have begun with the roads and conservation areas, or the suitable geology and roads, and so forth. Note something else: The GIS processes like reclassify or query are shown in a circle, while layers of data are shown in rectangles. Study this cartographic model for a moment you will be required to produce one for your semester project. (pause) Well talk more about this next time. Ok - lets summarize.
[Slide Summary ]
Today, we looked at data analysis operations. We saw how to calculate distances and areas in both vector and raster space. We saw how to form combination queries using Boolean operators. Proximity or buffer analyses tell us how close things are to an entity of interest. Overlay operations allow us to create new spatial data from two or more input data layers. Of all the GIS operations, overlay is probably the most important analysis function.
Interpolation methods are used to form surfaces. Once these surfaces are formed, especially DTM surfaces, we can use the slope and aspect analyses to form new surfaces. And in Network analysis, we looked at the solution of a shortest path problem.
And finally and most importantly, we saw how to construct a cartographic model which diagrams the plan we are going to use to solve a particular problem. (Next Slide Please)
[Slide Whats Nexpr~
6
J
y$78Y,F.Uij
!"8]pqj~il ; G #!?!B!'")"""Q#¸ªªªªªªªªªh|@56B*\]phh|@5B*phh|@B*phh|@56B*phh|@h|@56>*h|@>*CJaJh|@CJaJh|@5B*CJaJphC^_pqr
IJ
8Y,FG-.jd`d$a$=?"8qrPQJ #!_!"F"""@##
&F
dhd^h
&F
ddQ#W#_#a#~##%%%%L'`'c''(()****++++////2B2z244 5667E9Y9{9:":#:A:;;;;>>>8>AAAeCzCChDiDtDvDDdEeExEyEEh|@56B*phh|@56B*CJaJphh|@B*phh|@5B*CJaJphh|@h|@56B*phh|@B*phh|@56B*\]phC#%%c''()**++++//D2z24 56677[9{9#:A:;;d;>8>>?@AAzCCvDDDDZEdEyEEEEEFGpIhd^h
&Fd
&Fd
&Fdd^`dEEEEFZInIpIIHK\K~KLLLqNNNNR(RJRTTTVWW?@$h]ha$h]h&`#$d$&P1h/ =!"#$%ne8+~nU =PNG
IHDR}+WPLTEUbKGDHcmPPJCmp0712OmIDATXX˒0*t'Upul,mQ4Xb$H O(?L>I
t2d@YTm3 Í~ =O=5rM.dd[~GCܲ_
˂uM=sqJP׃&}J4ZU
A#8pDlJqh'<4ՑԆFfІrJA۠(6-3D,MRk6#(H[ҙ=V/m*jD*I~îrU{'.k"DrזJ 9YV+e3GC#FpX"l˲*5k݉ eiOVR|l
Qݲ}6W,nx8h8fzI3$ent l5(ĬJP]Zs`v_!d&h=k,۪f*B*
phFV@FFollowedHyperlink>*B*
phffH"HBalloon TextCJOJQJ^JaJ4 @24Footer
!.)@A.Page Number4@R4Header
!@@^_pqrIJ8Y ,FG-.j"8qrPQJ#_F@c !""####''D*z*, -..//[1{1#2A23368667899z;;v<<<<Z=d=y=====>?pAA\C~CDDFF(JJJLLOA00000000000000000000000000000000000000
0
0
0
0000000
00
00
000000000000000x0x0x0x00x0x0x0x0x0x0000x0x0x0x0x0x000x0x0x0x0x 0x 0x 0080x0x0x0x0x00x000x0@0x0x0x0x0x0x0x0x0x0x0x0x0x0x00x0x00x0x0x0x0x0x0x0x0x0x0x0x0x0x0x0x0x0x0x0x0x0x0x 0x 0x 0x 0x0x000x0x0x0x00x0x0x
{00.
{00.
{00.@0
{000<
;0&06669Q#EȈ@IMPS#;pIDo@JLNOQRU?K %039!!l,b$e8+~nU =&@0(
j
FA ?toptxtr@`#t_`#_`#P`#l`#`#T!`#!`#ĕ`##`#`#U((llw 7A
11v{{ !!7A
9*urn:schemas-microsoft-com:office:smarttagsplace=*urn:schemas-microsoft-com:office:smarttags PlaceType=*urn:schemas-microsoft-com:office:smarttags PlaceName8
*urn:schemas-microsoft-com:office:smarttagsCity
\\]]^^8o@oyyyy'+:<=>A ^a
47MP.*0*E*u*/
///52?233467699==LLWN'+:=>A3333333333333333333333#'_h<<<<N'+<=A<=ALeoneDonald LeonesaleonsaleonsaleonsaleonsaleonF#^X4#5}P6Vޱ'6tJ.Q$0_"9 22{}/4#?;@|3EjKN|6:[PweM[~}Au\Pfd}5iV%Z$I{bV~Eh
hh^h`hH.h
88^8`hH.h
L^`LhH.h
^ `hH.h
^`hH.h
xLx^x`LhH.h
HH^H`hH.h
^`hH.h
L^`LhH.h^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(n^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(n^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(n^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(n^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(n^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(n^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(nh^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(n^`OJPJQJ^J.^`.pp^p`.@@^@`.^`.^`.^`.^`.PP^P`.^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(nh^`OJQJo(hHnh^`OJQJ^Jo(hHohpp^p`OJQJo(hHh@@^@`OJQJo(hHh^`OJQJ^Jo(hHoh^`OJQJo(hHh^`OJQJo(hHh^`OJQJ^Jo(hHohPP^P`OJQJo(hH^`OJQJo(hHn^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(n^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(n^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(nh88^8`OJQJo(hHnh^`OJQJ^Jo(hHoh ^ `OJQJo(hHh^`OJQJo(hHhxx^x`OJQJ^Jo(hHohHH^H`OJQJo(hHh^`OJQJo(hHh^`OJQJ^Jo(hHoh^`OJQJo(hH^`OJQJo(n^`OJQJo(npp^p`OJQJo(n@@^@`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(n^`OJQJo(nPP^P`OJQJo(neM[{}/ $0'u\PfVF#$I{:[0}5i#?;KN3EJ#5} | l^nGl#85Pl22ع$<V.
1ązt$`pH\<Q^JBPzT
6\ċFyHU(J> ;4|XL0nNdBy:wLZB%&`ްi,nuNM݄u~x$Hĳ\Vr""Dz=T`SBfH"E4h4~Ȁ"pFL>̀bzBE.ҟqhG$.:誒2bG\PR'(px*
\&k.r T`SBfH"E4h4~Ȁ"pFL*p~6IJj%.fDJ4ZbFPZx]̎?@ABCDEFGHIJKLMNOPQRSTUVWYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root Entry FJM)<1TableXWWordDocumentSummaryInformation(DocumentSummaryInformation8CompObjj
FMicrosoft Word Document
MSWordDocWord.Document.89q