Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

- Topic stereoscopy, Parallax, Relie... by srinivas2036 5152 views
- Numerical from-photogrammetry-and-r... by Gokul Saud 296 views
- Lecture 1-aerial photogrammetry by virajain 27655 views
- Vertical aerial photographs by Collation Soft So... 1092 views
- Photogrammetry 1. by Naveen Kumar 16131 views
- Photogrammetry- Surveying by Gokul Saud 33600 views

No Downloads

Total views

7,342

On SlideShare

0

From Embeds

0

Number of Embeds

3

Shares

0

Downloads

422

Comments

0

Likes

7

No embeds

No notes for slide

- 1. GE 178 Lecture 6:Distortion and DisplacementRelief DisplacementTilt Displacement
- 2. DISTORTION VS. DISPLACEMENT
- 3. Distortionshift in the location of an object, which changes the perspective characteristics of the photo
- 4. Types of Distortion1. Film and Print Shrinkage – negligible effect*2. Atmospheric Refraction of Light Rays – negligible effect*3. Image Motion4. Lens Distortion*Except for precise mapping projects
- 5. Lens Distortion small effects due to the flaws in the optical components (lens) of camera systems leading to distortions typically more serious at the edges of photo radial from the principal point makes objects appear either closer to, or farther from the principal point than they actually are may be corrected using calibration curves examples: car windows/windshields, carnival mirrors
- 6. Lens Distortion Lens Distortion
- 7. Displacementshift in the location of an object in a photo, which does not change the perspective characteristics of the photo fiducial distance between an objects image and its true plan position, caused by change in elevation
- 8. Types of Displacement1. Curvature of the Earth – negligible effect*2. Relief Displacement – radial from the nadir3. Tilt Displacement – radial from the isocenter*Except for precise mapping projects
- 9. Major Causes of Non-uniformity inScale within a Single Photograph1. Relief Displacement2. Tilt Displacement
- 10. ReliefDisplacement
- 11. Relief Displacement Error in the position of the point in a photograph because of relief The position of a point in the photograph (which has a central projection) is different from its corresponding position on the map (which has an orthogonal projection) due to relief Radial from the nadir (assuming a vertical photograph, therefore, nadir = center of photo)
- 12. Relief Displacement
- 13. Relief DisplacementThe farther a point is from the nadir, the greater thedisplacement
- 14. Relief Displacement
- 15. Relief Displacement
- 16. Relief Displacement r’ CASE 1: ∆r Point is above the datum plane f Hmge (flying height) ∆h datum plane
- 17. Relief Displacement CASE 2: r’ Point is below the datum plane ∆r f Hmge (flying height) datum plane ∆h
- 18. Relief Displacement
- 19. a‘ a Class r Exercise: Dr Derive the equation for relief displacement DrA’A
- 20. Formula for Relief Displacement r h Dr HWhere: r’ = erroneous radial distance from the center of photo h = height/elevation of the point above/below the datum plane H = flying height above the datum plane
- 21. General Conclusion:Elevation and Relief DisplacementThe higher the point is above the datum plane (or the lower it is below the datum plane), the greater the relief displacementThe higher the flying height, the lesser the relief displacement r h Dr H
- 22. Corrected Radial DistanceIf the point on the ground is ABOVE the datum, the corrected position will be towards the center r r DrOtherwise, if the point is BELOW the datum, the corrected position will be away from the center r r Dr
- 23. Occlusion
- 24. Occlusion
- 25. How can we minimize ∆r? Use only the central part of the photograph (discard the edges) Fly higher but this would yield a smaller photoscale Fly higher, and use a camera with a larger focal length (for example, use a normal angle camera instead of a wide-angle camera) r h Dr H
- 26. ExampleA 1:15000 aerial photograph was taken using a wide-angle camera. A point on the photograph was identified and its measured distance from the center is 5.4 centimeters. If the corresponding point on the ground is elevated from the datum by 60 meters, determine the displacement due to relief and the correct radial distance of the point from the center of the photo.
- 27. Solutionf 6 inches 1 6* 2.54* 1100 r r Dr15000 H r 5.4 0.1417322H 2286 meters r 5.2582678 cms. r Dh (0.054)(60)Dr H 2286Dr 0.001417322 metersDr 0.1417322 cms.
- 28. Quiz 1 (1/4 Sheet of paper)The top and bottom of a utility pole in animage are 129.8 mm and 125.2 mm,respectively, from the principal point of avertical photograph. What is the height of thepole if the flying height above the base of thepole is 875m?
- 29. Tilt Displacement t a’b’’ p i a’’ n b’
- 30. Tilt Displacement An error in the position of a point on the photograph due to indeliberate tilting of the aircraft Due to instability of aircraft May be due to tilting of the aircraft along the flight line and/or perpendicular to the flight line Increases radially from the isocenter
- 31. Tilt Displacement t ∆tb a’ ybb’’ p i a’’ n b’ ya ∆ta
- 32. Principal Line Line of maximum tilt Line connecting the principal point, isocenter and nadir All lines perpendicular to this line are lines of zero inclination or zero phototilt this means that all points along a perpendicular line have uniform scale
- 33. Tilt Displacement t ∆tb a’ ybb’’ p i a’’ n b’ ya ∆ta
- 34. Phototilt (t) Amount of tilt of the aircraft (and thus the camera lens) with respect to the vertical axis Angle of tilt between the line perpendicular to the horizontal datum and the line perpendicular to the lens
- 35. Formula for Phototilt dS Sb S a sin t H mge H mge y yWhere: t = phototilt Sa = scale of first point, projected to the principal line Sb = scale of second point, projected to the principal line y = distance between a and b along the principal line Hmge = flying height with respect to the mean ground
- 36. Locating the Nadir and Isocenter Nadir – radial center of relief displacement Isocenter – radial center of tilt displacementdistance between p and n (pn) f tan tdistance between p and i (pi) f tan t 2
- 37. Formula for Tilt Displacement
- 38. Formula for Tilt Displacement y 2 sin t Dt f y sin tWhere: i = isocenter y = projection of erroneous radial distance from the isocenter (i) to the point along the principal line f = focal length t = phototilt
- 39. Corrected Radial Distance r r Dt if the point on the ground is above the horizontal r r Dt if the point on the ground is below the horizontal
- 40. Auxiliary Tilted Photo Coordinate System
- 41. Scale of a Tilted Photograph f y sin t S cos t H h
- 42. Tilt Displacement Practical SolutionPROBLEM: may cause large errors in determining scale and distancesSOLUTION: use 2 known or measurable ground distances that are: About the same elevation Equal distances from the photo center Diametrically opposite from the center
- 43. END OF LECTURE

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment