ge 119 photogrammetry 2 · lecture notes in ge 119: photogrammetry 2 topic 2. orientation of single...

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Topic 2. Orientation of Single Aerial

Photographs and Images

Division of Geodetic Engineering College of Engineering and Information Technology Caraga State University

GE 119 – PHOTOGRAMMETRY 2

Instructor: Engr. Jojene R. Santillan [email protected] [email protected]

Lecture Notes in GE 119: Photogrammetry 2 TOPIC 2. ORIENTATION OF SINGLE AERIAL PHOTOGRAPHS AND IMAGES 2

Outline

• Background and Principles behind the orientation of Aerial Photographs and Images

• Interior Orientation of Aerial Photographs and Images

• Exterior Orientation of Aerial Photographs and Images


Intended Learning Outcomes

• After this lecture and hands-on exercises, the students must have:

– Understood the mathematical concepts behind the orientation of single aerial photographs

– Performed orientation of single digital aerial photographs using digital photogrammetric software

– Explained the processes involved in the orientation of single aerial photographs.

4 Lecture Notes in GE 119: Photogrammetry 2 TOPIC 2. ORIENTATION OF SINGLE AERIAL PHOTOGRAPHS AND IMAGES

BACKGROUND AND PRINCIPLES BEHIND THE ORIENTATION OF AERIAL PHOTOGRAPHS AND IMAGES


Recall: A Typical Workflow in Photogrammetry

Generally: • Begins with the

capture of the images

• Calculation of the orientation parameters of all images that will be used

• Measure co-ordinates • Create several image

products • Use of results in

cartographic or GIS software

Source: Linder, W., 2016. Digital Photogrammetry – A Practical Course, Springer-Verlag Berlin Heidelberg, Germany.


The Orientation Process in Photogrammetry

Orientation the process of

establishing the relation between two coordinate systems of an aerial photograph or image


Recall: Geometry of a Vertical Aerial Photograph

Aerial Photograph

Source: Lillesand et al 2004, Remote Sensing and Image Interpretation Fifth Edition, Wiiley.


Photo-coordinate System = “Camera-internal coordinate System”

Aerial Photograph

• Each point in the aerial photograph can be located through the use of the “photo-coordinate system”

• This “photo-coordinate system” is defined based on the geometric properties of the camera system

• The photo-coordinate system’s origin is defined by the use of fiducial marks

• This coordinate system can be referred to as “camera-internal coordinate system”


Camera Internal Coordinate System

Aerial Photograph

x

y

Fiducial marks p

Object p’s location in the photo can be defined as (xp, yp)


Image Coordinate System of Digital Aerial Photographs or Images

• Aside from the “camera-internal coordinate system”, objects can also be located based on their pixel row and column numbers in a digital aerial photograph or image

• The pixel row and column numbers refers to the “image coordinate system” or more appropriately, “pixel coordinate system” denoted by the X‟ and Y‟ axes.

Aerial Photograph

x

y

p

X'

Y‟


Recall: Digital Image Characteristics

• Digital Image = a matrix of Digital Number (DN) values

• Each DN is located at a specific row and column in the matrix

• Pixel: - the smallest unit of an

image. - image pixels are normally

square and represent a certain area on an image (e.g., the ground segment within the sensor’s IFOV)

- Each pixel has DN value associated with it.


Image/Raster/Pixel Coordinate System

• x‟-axis = column number

• y‟-axis = row number; also called “line number”

• A pixel’s location can be identified by its (column, row) value

• The (1,1) axis is on the upper leftmost part of the images

• Cell height and cell width = depends on the image’s spatial resolution


In addition to (x,y) and (X‟, Y‟) coordinates…

• Objects in aerial photograph also have actual ground coordinates (X,Y, Z)!

• That means an object can be located in an aerial photo or image in terms of 3 coordinate systems!

Aerial Photograph

x

y

p

X'

Y‟

X

Y


• Ground coordinate system in 3D

X

Y

Z



• In aerial photogrammetry, we are interested in getting 3D spatial location of objects in aerial photographs in terms of their true ground coordinates

• The question now is, how do we obtain the 3D ground coordinates of objects from the aerial photograph?

• Or…How will we get the „true ground coordinates” from the aerial photograph given that we know the “camera-internal coordinates” of the objects?


• The answer: we need to find the relation between the three (3) coordinate systems (CS):

• The camera-internal coordinate system

• The pixel coordinate system

• The ground coordinate system

To find the relations, we apply the process of “ORIENTATION”

Camera-Internal CS

Pixel CS Ground CS


For single aerial photographs, there are 2 kinds of orientation procedures:

• Interior Orientation – establishing the relation between the

camera-internal co-ordinate system and the pixel co-ordinate system

• Exterior Orientation – establishing the relation between the pixel

co-ordinate system and the ground coordinate system


INTERIOR ORIENTATION OF AERIAL PHOTOGRAPHS AND IMAGES


Interior Orientation

• establishing the relation between the camera-internal co-ordinate system and the pixel co-ordinate system

Camera-Internal CS

x

y

p

Pixel CS

p

X'

Y‟


Output of Interior Orientation

• After doing an interior orientation, we would be able to determine the following:

– The pixel coordinates of each and all objects in the photograph/image

– The size of each pixel in the photograph in terms of “mm”.

– The dimension of the photograph/image in terms of number of rows and columns

But how do we do an “Interior Orientation”?


Consider this aerial photograph:


Aerial photographs are provided along with the following information: 1. Fiducial marks 2. Camera-internal coordinates of each fiducial marks (usually

provided in the camera calibration certificate)

FIDUCIAL MARKS 1

2

3

4


The camera-internal coordinates of each fiducial marks are known based on film format size and the calibration of the camera used.

1

2

3

4 FM4 = (0 mm, 113 mm)

FM1 = (113 mm, 0 mm)

FM2 = (0 mm, -113 mm)

FM3= (-113 mm, 0 mm)


To transform the camera-internal coordinates into pixel coordinates, a

two-dimensional (2D) transformation method called “AFFINE

TRANSFORMATION” is used.


Affine Transformation Equation

• If (x,y) is the camera-internal coordinates of an object in an aerial photograph, we wish to find its corresponding pixel coordinates (X’, Y’)

• Using Affine Transformation, the pixel coordinates can be determined as:

X‟ = a0 + a1x + a2y

Y‟ = b0 + b1x + b2y

• Note: a0, a1, a2, b0, b1 and b2 are called the „affine transformation parameters‟


Affine Transformation

• Using Affine Transformation, the pixel coordinates can be determined as:

X‟ = a0 + a1x + a2y

Y‟ = b0 + b1x + b2y

• How do we determine the values of the „affine transformation parameters‟?


We can relate (X‟, Y‟) to (x, y) by:

• Using the camera-internal coordinates of the fiducial marks as the (x,y) values

• How about (X‟, Y‟)? – Using a computer software, we need to find the

fiducial marks in the digital aerial photograph and determine its row and column values (i.e., their X’ and Y’ coordinates)

• To find the parameter values, a minimum of three (3) fiducial marks are required


We can then create a table like this:

Fiducial Mark

Camera-Internal Coordinates

Pixel Coordinates

x (mm)

y (mm)

X Y

1 113 0 2699 1363

2 0 -113 1369 2699

3 -113 0 28 1373

4 0 113 1358 37


We can then set-up the affine transformation equations: • For X’s:

– FM1: 2699 = a0 + a1(113) + a2(0)

– FM2: 1369 = a0 + a1(0) + a2(-113)

– FM3: 28 = a0 + a1(-113) + a2(0)

– FM4: 1358 = a0 + a1(0) + a2(113)

• For Y’s:

– FM1: 1363 = b0 + b1(113) + b2(0)

– FM2: 2699 = b0 + b1(0) + b2(-113)

– FM3: 1373 = b0 + b1(-113) + b2(0)

– FM4: 37 = b0 + b1(0) + b2(113)


Updating the equations:

• For X’s:

– FM1: 2699 = a0 + 113a1

– FM2: 1369 = a0 - 113a2

– FM3: 28 = a0 - 113a1

– FM4: 1358 = a0 + 113a2

• For Y’s:

– FM1: 1363 = b0 + 113b1

– FM2: 2699 = b0 - 113b2

– FM3: 1373 = b0 + 113b1

– FM4: 37 = b0 + 113b2

• If we use only 3 FM, we can easily compute for the values of a0, a1, a2, b0, b1, and b2

(i.e., number of equations = number

of unknowns)


• For X’s:

– FM1: 2699 = a0 + 113a1

– FM2: 1369 = a0 - 113a2

– FM3: 28 = a0 - 113a1

– FM4: 1358 = a0 + 113a2

• For Y’s:

– FM1: 1363 = b0 + 113b1

– FM2: 2699 = b0 - 113b2

– FM3: 1373 = b0 + 113b1

– FM4: 37 = b0 + 113b2

• If we use all the FMs, number of equations > number of unknown (redundancy of 2

equations)

• To determine the parameter values, LEAST SQUARES ESTIMATION IS USED!


After estimating the best values of a0, a1, … • The transformation equations:

X‟ = a0 + a1x + a2y

Y‟ = b0 + b1x + b2y

are used to easily convert (x,y) to (X‟, Y‟), and vice versa.


EXTERIOR ORIENTATION OF AERIAL PHOTOGRAPHS AND IMAGES


• establishing the relation between the pixel (or image) co-ordinate system and the ground coordinate system

X

Y

Z

Exterior Orientation



Exterior Orientation

• When we do exterior orientation, we define the geometric relationship between an object and its image.

• Once we have done exterior orientation, it becomes possible to reconstruct the spatial position of an object from its image.

• Exterior orientation is based on the condition of collinearity


The Collinearity Condition

• Collinearity is the condition in which:

– the exposure station of any photograph (or image),

– any object point in the ground coordinate system, and

– its photographic image

all lie on a straight line.



• Collinearity is the condition in which:

– the exposure station (L) of any photograph (or image),

– any object point (P) in the ground coordinate system, and

– its photographic image (p)

all lie on a straight line.

Ground

Photo/Image




• The collinearity condition holds irrespective of the angular tilt of the photograph.

• The possible angular rotations from that of an equivalent vertical photograph are:

– ω (“pitch”)

– φ (“roll”)

– К (“yaw”)

Ground

Photo/Image



ω, φ, К ….

Source: http://howthingsfly.si.edu/sites/default/files/image-large/Roll,-Yaw,-Pitch_lg_0.jpg

Source: http://doc.aldebaran.com/2-1/_images/rollPitchYaw.png

Source: https://www.novatel.com/assets/Web-Phase-2-2012/Solution-Pages/AttitudePlane.png


The Collinearity Equations

• The collinearity equations are the equations that express the collinearity condition

• They describe the relationships among pixel/image coordinates, ground coordinates, the exposure station position, and angular orientation of a photograph/image.



• The equations are non-linear

• Contain 9 unknowns:

– The exposure station position (XL, YL, ZL)

– The three rotation angles (ω, φ, К) which are embedded in the m coefficients

– The object point coordinates (XP,YP,ZP)



• The parameters (XL, YL, ZL) and (ω, φ, К) are commonly

referred to as the “Exterior Orientation Parameters”



• The “m” coefficients are elements of a 3x3 rotation matrix

cos φ cos κ −cos φ sin κ sin φ cos ω sin κ + sin ω sin φ cos κ cos ω cos κ − sin ω sin φ sin κ −sin ω cos φ sin ω sin κ − cos ω sin φ cos κ sin ω cos κ + cos ω sin φ sin κ cos ω cos φ

m =

• m is obtained by getting the product of matrices of rotation around the first (ω), second (φ) and third (κ) axis (i.e., m = R3*R2*R1)


m values

• m11 = cos φ cos κ • m21 = cos ω sin κ + sin ω sin φ cos κ • m31 = sin ω sin κ − cos ω sin φ cos κ • m12 = −cos φ sin κ • m22 = cos ω cos κ − sin ω sin φ sin κ • m32 = sin ω cos κ + cos ω sin φ sin κ • m13 = sin φ • m23 = −sin ω cos φ • m33 = cos ω cos φ

cos φ cos κ −cos φ sin κ sin φ cos ω sin κ + sin ω sin φ cos κ cos ω cos κ − sin ω sin φ sin κ −sin ω cos φ sin ω sin κ − cos ω sin φ cos κ sin ω cos κ + cos ω sin φ sin κ cos ω cos φ

m = cos φ cos κ −cos φ sin κ sin φ cos ω sin κ + sin ω sin φ cos κ cos ω cos κ − sin ω sin φ sin κ −sin ω cos φ sin ω sin κ − cos ω sin φ cos κ sin ω cos κ + cos ω sin φ sin κ cos ω cos φ



• The collinearity equations are “at the heart” of softcopy (e.g, digital) photogrammetric operations

• If the location of the exposure station is known as well as the angular rotations, then any position on the ground can be located in the photo or image.


• Since there are 9 unknowns, how do we proceed with the exterior orientation?


Solving the Collinearity Equations to do Exterion Orientation

• 1st Approach:

– Use of combined GPS/GNSS and IMU during photo/image acquisition to determine the 6 exterior orientation parameters

• GPS/GNNS to determine (XL, YL, ZL)

• IMU or Inertial Measuring Unit to determine (ω, φ, К)

Source: http://www.mdpi.com/remotesensing/remotesensing-04-01519/article_deploy/html/images/remotesensing-04-01519f1-1024.png


Example Set-up to get the 6 orientation parameters using GPS and IMU

Source: http://www.mdpi.com/remotesensing/remotesensing-04-01519/article_deploy/html/images/remotesensing-04-01519f1-1024.png


Solving the Collinearity Equations to do Exterior Orientation

• 2nd Approach:

– Through the process of “space resection”


Space resection (or photo-resection)

• Similar to the “resection” method in Surveying

• Use of Ground Control Points (GCPs)

• In this process, known Ground Control Points are located in the image, and the image coordinates are determined

• This means, for each GCP we know the following variables of the collinearity equations: – XP, YP, ZP

– xp, yp

• We are left with the following unknowns: – XL, YL, ZL

– ω, φ, К (or the rotation matrix coefficients)


Space resection (or photo-resection)

• To determine the unknowns, we need to have at least three (3) GCPs to set-up 6 equations which can be solved simultaneously

– More than 3 GCPs are recommended when doing exterior orientation to allow redundancy and to get the best estimates of the exterior orientation parameters

– If more than 3 GCPs, then more than 6 equations can be formed and a least squares solutions for the unknown is performed • Digital photogrammetric software can do this

automatically


Using GCPs for Exterior Orientation

• In photogrammetry, a GCP is an object point which is represented in the images and from which the three-dimensional object (ground) coordinates (XP, YP, ZP) are known



• Using GCPs in exterior orientation means:

– We have to look for points in our image

– Then:

• Find these points for instance in a topographic map and get their coordinates out of the map (e.g., by manually measuring XP and YP, and interpolating the elevation between neighboring contours to get ZP)

• Or: Conduct GPS/GNSS survey to obtain the 3D coordinates of the GCPs



• For each image, we need at least 3 well-distributed GCPs

– Well-distributed means that the 3 GCPs should form a triangle, not a line

• Basic rule: “The more, the better” to get a stable over-determination of the exterior orientation parameters

– Will allow error checking (e.g., how good our selected GCPs are)


Two Kinds of GCPs used during Exterior Orientation

• Signalized (targeted) GCPs: – These GCPs are set-up on the

ground before taking the photos

– Example: • GCPs “signalized” using white

bars forming a cross with the point itself marked with a central “dot” or “square” – The bars are usually 1.2 x 0.2

m in size

– The dot is usually 0.2 m diameter

– However, dimensions will depend on the photo scale

Source: https://cdn-images-1.medium.com/max/1600/0*3FhWS4peJNBdupE4.

Source: http://old.grida.no/images/thumbs/3.png


Example of Signalized GCPs


Another Example of Signalized GCPs

Source: https://pbs.twimg.com/media/C7rEUfwVMAEnvg-.jpg


Appearance of a “Signalized” GCP in an Image

Source: http://slideplayer.com/slide/2520727/9/images/15/GCP+Measurement+Signalized+GCP+/+ICP.jpg


Two Kinds of GCPs used during Exterior Orientation

• Natural GCPs – Real object (ground/terrain) points which we can

clearly identify in the image as well as in a topographic map (or in the field when we do GPS/GNSS survey)

– More commonly Used – Suggested Natural GCPs:

• Objects with rectangle corners (e.g., buildings) • Small circle-shaped points

– Important: • Natural GCPs must have 3D coordinates:

– Do not select GCPs whose elevation is difficult or not possible to determine (from a topo map or from GPS/GNSS survey)

» Example: Rooftops

• Prefer selecting GCPs that are on the ground.


Examples of Natural GCPs

Source: Linder, W., 2016. Digital Photogrammetry – A Practical Course, Springer-Verlag Berlin Heidelberg, Germany.


Example GCP Data

Indicative Location of the GCPs in the photo


Remarks

• To experience how interior and exterior orientation works, a lab exercise will be given in the coming weeks (after the Prelim Exam)


Further Reading

• Linder, W., 2016. Digital Photogrammetry – A Practical Course, Springer-Verlag Berlin Heidelberg, Germany.

• Lillesand et al 2004, Remote Sensing and Image Interpretation Fifth Edition, Wiiley.

(See Facebook page for the PDF link)

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