Name:__Key________________

GIS in Water Resources Midterm Exam                                                                 Fall 2006

There are 4 questions on this exam. Please do all 4.

1.  [20 points]  Geodesy, Map Projections and Coordinate Systems

Below are the spatial reference properties of the DEM used for hydrologic analysis in Exercise 4.

a)      What horizontal earth datum is used?

North American Datum of 1983

b)      What map projection is used? Why is this particular projection used for this exercise?

Albers equal area, it preserves earth surface area which is important for mass balances

c)      What are the geographic coordinates of the origin (φo, λo)?

(18,00° -100.00°)

d)      What are the projected coordinates of the origin (Xo, Yo)?

(1500000, 6000000)
2.
[25 points]  Distances on a Curved Earth

Logan, Utah is located at 41°44'N, 111°50'W.

Austin, Texas is located at 30°11'N, 97°40'W.

a) Convert these coordinates to decimal degrees and indicate which of these numbers represents longitude and which represents latitude by filling the corresponding decimal degree longitude and latitude into the following table

 Cities Longitude Latitude Logan -(111+50/60) =  -111.8333o 41+44/60  =  41.7333o Austin -(97+40/60) =    -97.6667o 30+11/60  =  30.1833o

b) Assume a spherical earth with radius of 6370 km. Calculate the distance from Logan (Utah)  to Austin (Texas).

f1, f2, are the latitudes, 41.7333ºand 30.1833º respetively

l1, l 2, are the longitudes.  l 1 - l 2 = -111.8333- (-97.6667) = -14.1666º

Now Cos-1(0.9602) = 16.22199 º but we need the result in radians, so 16.22199*π/180 = 0.2831 radians, so

Evaluating the formula d=1803 km

c) Discuss some other ways that you have learned for calculating the distance from Logan to Austin that are not limited to assuming a spherical earth and describe how you would go about calculating this distance more precisely.  (What we are looking for here is a description of how you would do this, using the GIS knowledge and tools you have learned.  You are not expected to do it in this question.)

One way is to project the locations into a GIS spatial reference, preferably chosen to minimize distance distortions then calculating the distance based on projected coordinates.  This does however involve some level of distortion associated with the projection.  A more precise approach is to use the ellipsoid that underlies the horizontal datum used to define the coordinates.  The distances on a curved earth handout provided a matlab script that implemented ellipsoid calculations.

3.  [30 points]  Hydrologic Variables derived from DEM’s

Following is a grid of elevations.  Because in general it is not possible to unambiguously determine flow directions around the edges, these have been specified for you as indicated.

a)      On the above grid, determine which grid cells are pits and indicate the elevation to which they need to be raised to fill them.

b)      Determine the flow direction grid using the 8-direction pour point method (D8) for the 9 internal grid cells.  Indicate the flow direction by using an arrow in each cell on the grid below.

c)      Determine the flow accumulation grid corresponding to the D8 flow directions.  Label each cell on the grid below with the number of upstream cells draining into it (ESRI convention).

d)      On the above flow accumulation grid, map the streams corresponding to a flow accumulation threshold greater than or equal to 5.

e)      Assume that these are 30m grid cells and that the mean annual rainfall over this area is 750mm.  Consider that on grid cells that are streams (as determined by the theshold of 5 cells above) all the rainfall becomes runoff, but that on grid cells that are not streams 80% of rainfall is lost through evapotranspiration and infiltration and only 20% of rainfall becomes runoff.  Determine mean annual flow into grid cell A in m3/year.

 There are 11 grid cells that drain in to A.  9 (light shading) are not on the stream.  2 (dark shading) are on the stream.  From the on-stream cells flow is: 2 * 750 mm * 900 m2 /1000 mm/m = 1350 m3 From the off-stream cells flow is: 9 x 750 mm * 0.2 * 900 m2 / 1000 mm/m = 1215 m3 Combining, the flow is: 2565 m3/year

4.  [25 points]  ArcHydro Catchments and Networks.

Following are subsets of the “Catchment”, “DrainageLine” and “Drainage Point” feature classes derived using Arc Hydro.

 14

 19

 135

 140

 138

 131

 128

 26

 13

 10

 416

 413

 412

 424

 423

The drainage points have each been labeled with their HydroID.  Following are the corresponding attribute tables.  The units are km for length and km2 for area.

Drainage Point

 HydroID DrainID 1 412 10 2 413 14 3 416 19 4 423 13 5 424 26

DrainageLine

 Shape_Length HydroID NextDownID DrainID 1 57 128 135 10 2 26 131 135 14 3 17 135 140 13 4 31 138 140 26 5 12 140 156 19

Catchment

 OID Shape_Length Shape_Area HydroID NextDownID 1 162 437 10 13 2 99 177 14 13 3 52 69 19 18 4 76 94 13 19 5 46 392 26 19

a)      On the map above (on the previous page) label each DrainageLine and Catchment with its HydroID

DrainageLine HydroIDs are labeled in red text with black border.  Catchment HydroIDs are black text with no border.

b)      Report the area of the Catchment that drains to the point with HydroID 423.

HydroID 423 in the drainage line table has DrainID = 13 which points to the HydroID field in the catchments table.  For the catchment with HydroID = 13,  A = 94 km2

c)      Evaluate the total length of drainage lines draining to the point with HydroID 423

Length HydroID-135 +  Length HydroID-131 + Length HydroID-128

L = 17 + 26 +57 = 100 km

d)      Evaluate the total area of the watershed draining to the point with HydroID 423.

Area Hydro ID-10 +  Area Hydro ID-14 + Area Hydro ID-13

A = 437 + 177 + 94 = 708 km2

e)      Evaluate the drainage density of the watershed draining to the point with HydroID 423

D = Length/ Area = 100/708 = 0.141 km/ km2