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Image 1 and 2 form a stereopair.

- Object A on the ground in Image 1 has a photo-coordinate from its principal point of \([x = 1.7 \, \text{mm}, y = 2.1 \, \text{mm}]\).
- Object A on the ground in Image 2 has a photo-coordinate of \([x = -2.1 \, \text{mm}, y = 1.2 \, \text{mm}]\).

Assume truly vertical photos acquired with:
- Camera focal length \(f = 16 \, \text{mm}\)
- Flying height of 600 m above MSL
- Baseline separation of 75 m

Use the parallax equations from Chapter 3 to answer the questions below:

(a) Measure horizontal parallax to compute the height \(h\) in feet of Object A above the MSL datum.
*Note: Use 0.3084 meters per foot for unit conversions.*

(b) The exposure station location "L" for Photo 1 (see example Figure 3.16 in text) has a ground principal point coordinate in NAD83 State Plane Texas South (US Survey Feet) of 1,341,037.5 ft E, 17,181,413.4 ft N. What is the easting and northing coordinate in survey feet for Object A? Show work.
*Assume Photo 1's photo-coordinate system (y-axis) is oriented to true north. Horizontal coordinates should be carried out to 0.01 ft.*

Please refer to the stereo-pair image showing conjugate principal points of Photo 1 and Photo 2 for the baseline separation and overlap area.

Answer :

Answer:

The height of Object A above the MSL datum is approximately 0.2506 feet. The easting coordinate of Object A is 1,341,037.5 ft, and the northing coordinate of Object A is approximately 17,180,765.75336 ft.

Explanation:

To calculate the height of Object A above the MSL datum, we need to measure the horizontal parallax. The horizontal parallax is the difference in the x-coordinates of Object A in Image 1 and Image 2.

Given:

  • Photo-coordinate of Object A in Image 1: [x = 1.7 mm, y = 2.1 mm]
  • Photo-coordinate of Object A in Image 2: [x = -2.1 mm, y = 1.2 mm]
  • Camera focal length (f): 16 mm
  • Flying height above MSL: 600 m
  • Baseline separation (B): 75 m

Using the parallax equations, we can calculate the horizontal parallax:

Horizontal Parallax = (x2 - x1) * (f / B)

Substituting the values:

Horizontal Parallax = (-2.1 mm - 1.7 mm) * (16 mm / 75 m)

Horizontal Parallax = -3.8 mm * (16 mm / 75 m)

Horizontal Parallax = -0.812 mm

Now, we can convert the horizontal parallax to feet using the conversion factor of 0.3084 meters per foot:

Horizontal Parallax (in feet) = -0.812 mm * (0.3084 m/ft)

Horizontal Parallax (in feet) = -0.2506 ft

Therefore, the height of Object A above the MSL datum is approximately 0.2506 feet.

To find the easting and northing coordinates of Object A, we need to use the ground principal point coordinates of the exposure station for Photo 1 and the photo-coordinates of Object A in Image 1.

Given:

  • Ground principal point coordinate for Photo 1: [Easting = 1,341,037.5 ft, Northing = 17,181,413.4 ft]
  • Photo-coordinate of Object A in Image 1: [x = 1.7 mm, y = 2.1 mm]

Since the photo-coordinate system (y-axis) is oriented to true north, the easting coordinate of Object A will be the same as the easting coordinate of the exposure station.

Therefore, the easting coordinate of Object A is 1,341,037.5 ft.

To find the northing coordinate of Object A, we need to subtract the y-coordinate of Object A in Image 1 from the northing coordinate of the exposure station:

Northing Coordinate of Object A = Northing Coordinate of Exposure Station - y-coordinate of Object A in Image 1

Northing Coordinate of Object A = 17,181,413.4 ft - 2.1 mm * (0.3084 m/ft)

Northing Coordinate of Object A = 17,181,413.4 ft - 0.64764 ft

Northing Coordinate of Object A = 17,180,765.75336 ft

Therefore, the northing coordinate of Object A is approximately 17,180,765.75336 ft.

Learn more about parallax equations here:

https://brainly.com/question/28680890

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