Lab 7

Introduction

The goal of this lab is to develop skills in performing photogrammetric tasks on aerial photographs and satellite images.   Specifically the lab is designed to train and understand the mathematics behind the calculation of photographic scales, measurement of areas and perimeters, and calculating relief displacement. 

Methods 

Part 1: Scales, measurements and relief displacement. 

From the image above the task was to create a scale of the photograph by measuring the distance between point A and B.  The real world distance is 8822.47 feet and by measuring the distance with a ruler was 2.25 inches or .1875 ft.  From here both measurements were divided by .1875 feet to give the scale of 1/47,053.17.  

The second part of the question also has to do with scale.  This time with the aerial photograph, ec_east-sw.img.  The photograph was acquired at a height of 20,000 feet above sea level with a camera focal length lens of 152 mm.  The elevation of Eau Claire County is 796 feet.   Using an equation Scale = f/ H-h the answer can be seen in the results tab below. 

Section 2: Measurement of areas on aerial photographs 

In Erdas Imagine open the image ec_west-se.  Click measure>show panel>point drop down arrow> and polygon tool.  Then the figure near x was digitized to find the area. To complete the polygon double click on the last point to complete the circle.  For this part we want the area in hectares and acres, it can be switched near the top right below manage data.  

The same steps were taken to find the perimeter of the feature but instead of choosing polygon, polyline was chosen.  For this step we want the perimeter in meters and miles. 

The figure below is an example of digitizing the lake to give either the area or perimeter. 

Section 3: Calculating relief displacement from object height. 

Using the image ec_west-se the relief displacement of the smoke stack, A, was determined from the image below.  The scale of the image is 1:3209 and the height of aerial photograph is 3980 feet.  The smoke stack measured at .4 inches and it is 8.7 inches away from the principal point.  Real word of smoke stack =1283.6 inches.  Equation then 1283.6 x 8.7/47, 760 inches.  3980 feet = 47,760 inches 

Part 2: Stereoscopy 

In Erdas Imagine open the image ec_city and in a second view bring in the image ec_dem2.  Then click the Terrain-Anaglyph button to open the Anaglyph geration window show below.  

Anaglyph Generation Window 

For Input DEM, input ec_dem2 and input image use ec_city.  Then click on the folder for output image and name it ec_anaglyph.  Increase the vertical exaggeration to 2  and all the other factors should be left at default.  Dismiss the run and bring in the new image for results, which can be seen below.

Part 3: Orthorectification 

The final result should give you something similar to the image below,

Results

Part 1: Scales, measurements and relief displacement. 

   1.      A-B= 2.25 inches, =.1875 ft.  .1875/8,822.47=

1/47,053.17   ft.

    2.      152mm/20,000-796 ft.  152mm = 15.2 cm.  15.2cm=5.98 in.  5.98in = .498 ft.

.498/19,204 ft = 1/38,562.249  ft. 

Section 2: Measurement of areas on aerial photographs 

a.  38.83 hectares   95.97 acres

b. 4,122.07 meters   2.56 miles

Section 3: Calculating relief displacement from object height. 

        0.4 inches of smoke stack,  8.7 inches to principal point.  .4 x 3209 for scale = 1283.6 inches.  Equation = 1283.6 x 8.7/ 47,760 inches.  Relief Displacement = 0.23 inches.

Because the object is above the datum and leaning outwards, the objects and features must be plotted inwards. 

Part 2: Stereoscopy 

Describe the elevation of features in Eau Claire.

The elevation features in Eau Claire appear very flat, probably because Eau Claire does not have much elevation change.  You can easily see elevation change around the Chippewa River and some lakes.  It is hard to see elevation change in the city. 

How different are these features from reality? 

 The features to me, do not appear to differ in elevation as much as they would in reality.  When the image appears in the same color it is hard to tell a difference in elevation especially in the city where it appears grey.  I cannot see a big difference in elevation near “the hill” on campus or on State Street. 

What factor(s) might be responsible for differences between what you observe in the city and what you are now seeing in the anaglyph image?

One factor is that the image on the computer screen is hard to view in 3D, but in real life everything is 3D.   Also the scale of the image and spatial resolution do not contribute to the easiness to view the images elevation change.

Part 3: Orthorectification 

Comment on the degree of accuracy of spatial overlap at the boundaries of the two 

orthorectified images.

 The spatial accuracy of the overlap between the two images is very clean.  The only thing that stands in the way of the two images is the black line that separates between the images.  When zooming in it is the cleanest line between two images I have ever seen in during this class. 

Sources

All images were provided by Dr. Cyril Wilson of the geography department at the University of Wisconsin Eau-Claire.  The course is title Geogrpahy 338: Remote Sensing of the Environment