5/24/2015 GEM 3366 1 Lecture 7 Content Photo / Map Projection Definition of common terms Errors on...
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Transcript of 5/24/2015 GEM 3366 1 Lecture 7 Content Photo / Map Projection Definition of common terms Errors on...
04/18/23GEM 3366
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Lecture 7 Content
Photo / Map Projection
Definition of common terms
Errors on photo
Mathematical relationship (Scale)
Vertical photos
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Photo / Map Projection
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Definition of common terms
ө
n
o
N
p
P
Plane II
Plane I
R
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– ““Plane I” represents the plane taken by the Plane I” represents the plane taken by the negative / photo which is called negative / photo which is called image image planeplane
– “ “Plane II” represents the plane taken by Plane II” represents the plane taken by the terrain which is called the terrain which is called object planeobject plane
– the two planes intersect at a line called the the two planes intersect at a line called the horizontal tracehorizontal trace
– ““o” is the o” is the optical centeroptical center of the camera lens of the camera lens
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– ““p” is the p” is the principal pointprincipal point on the image plane, on the image plane, defined as the foot of the perpendicular from “o” defined as the foot of the perpendicular from “o” to “Plane I”to “Plane I”
– ““o” is the o” is the optical centeroptical center of the lens of the lens– ““P” is the P” is the ground principal pointground principal point– ““n” is the n” is the nadirnadir point point, which is the photo plumb , which is the photo plumb
pointpoint– ““N” is the N” is the ground plumb pointground plumb point from the nadir from the nadir
pointpoint– “ “pR” is the principal line, “po” is the Principal pR” is the principal line, “po” is the Principal
distance of focal length. “opR” are on the distance of focal length. “opR” are on the Principal PlanePrincipal Plane
– ““өө” is the angle of tilt” is the angle of tilt
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Mathematical relationships from a vertical photoScale:
f
H
Terrain
Photo
1/S = f/H where S = Scale Factor
LEARN
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Isocenter and Nadir Points – Principal Point is the
point where a perpendicular projected the center of the lens intersects the photo image
– Nadir Point is the point vertically beneath the camera lens at the time of exposure
– Isocenter is the point on the photo that falls on a line approximately halfway between the principal point and the nadir point
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Isocenter and Nadir Points
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Equations on the Principal Plane
of
n
i
p
N I P
ө
ө
R
Note: Isocenter (i) is the point in which the bisector of the angle between the plumb line and the principal axis meets the image plane. v
Image plane
Object planeH
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1. angle nop = ө and angle poi = angle ion = ө/2
2. Distance from principal point to plumb point
pn = po tan ө = f tan ө
3. Distance from principal point to isocenter
pi = po tan ө/2 = f tan ө/2
4. Distance from principal point to horizon
pv = po cot ө = f cot ө
5. PN = No tan ө = H tan ө
PI = PN – NI = H tan ө - H tan ө/2 = H (tan ө - tan ө/2)
Homework:
Prove the following using the previous diagram
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Vertical photos – These are photos resulting if the axis of These are photos resulting if the axis of
the camera is exactly vertical when the camera is exactly vertical when exposure is madeexposure is made
– Despite all precautions, small tilts, Despite all precautions, small tilts, generally less than 1generally less than 1˚ and rarely greater ˚ and rarely greater than 3˚, are invariably presentthan 3˚, are invariably present
– ScaleScale is determined by: is determined by:Ratio of a distance on a map to that same Ratio of a distance on a map to that same length on the ground,length on the ground, or or
Ratio of the focal length to that of the flying Ratio of the focal length to that of the flying height for a flat area or the average height of a height for a flat area or the average height of a hilly areahilly area
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Vertical photos
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– Using similar triangles:Using similar triangles:
LabLab and and LABLABLA
La
AB
ab
– Using similar triangles:Using similar triangles:
LoaLoa and and LOALOA
AhH
f
LA
La
– Equating the two equations and recognizing Equating the two equations and recognizing that equals photo scale at A and B and that equals photo scale at A and B and recognizing that AB is recognizing that AB is infinitesimally infinitesimally
short then the equation for the scale at A is:short then the equation for the scale at A is:
AB
ab
AA hH
fS
– Similar equations are generated for points B, Similar equations are generated for points B, C, and DC, and D
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An example:An example: A vertical photograph was generated using a camera with a A vertical photograph was generated using a camera with a
focal length of 6in at a flying height of 10,000ft. Above mean focal length of 6in at a flying height of 10,000ft. Above mean sea level. What is the photo scale at point ‘a’ if the elevation sea level. What is the photo scale at point ‘a’ if the elevation of point A on the ground is 2500ft above mean sea levelof point A on the ground is 2500ft above mean sea level
Solution:Solution:
000,15:11250
1
2500000,10
6
in
inin
hH
fS
AA
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Ground coordinates from a single vertical photograph
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– photo coordinates are xa, ya and xb, yb
– ground coordinates are XA, YA and XB, YB
AXax
AhHf
AAOao
– Using similar triangles:Using similar triangles:
LOLOAAAA΄́ and Loa and Loa΄́
ThenThen
f
xaAhH
AX
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AYay
AhHf
AAaa
– Using similar triangles:Using similar triangles:
LALA΄́A and LaA and La΄́aa
ThenThen
f
yaAhH
AY
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An example:An example: A vertical photograph was taken with a 152.4mm focal length A vertical photograph was taken with a 152.4mm focal length
camera from a flying height of 1385m above datum. Images camera from a flying height of 1385m above datum. Images ‘a’ and ‘b’ of two ground points A and B appear on the ‘a’ and ‘b’ of two ground points A and B appear on the photograph, and their measured photo coordinates are xphotograph, and their measured photo coordinates are xaa= -= -
52.35mm, y52.35mm, yaa= -48.27mm, x= -48.27mm, xbb= 40.64mm, and y= 40.64mm, and ybb= 43.88mm. = 43.88mm.
Determine the horizontal length of line AB if the elevations of Determine the horizontal length of line AB if the elevations of points A and B are 204m and 148m above datum, points A and B are 204m and 148m above datum, respectivelyrespectively
Solution for Solution for
point A:point A:
mY
mX
A
A
1.374)2041385(4.152
27.48
7.405)2041385(4.152
35.52
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… The End …