Ch5

‫داﻧﺸﮕﺎﻩ اﻣﺎم ﺣﺴﻴﻦ )ع(‬

‫ﻓﺼﻞ ‪:۵‬‬ ‫ﻋﻨﻮان درس ‪:‬‬

‫ﻣﻌﺎدﻻت رﻳﺎﺿﯽ ﺗﺤﺖ ﺷﺮاﻳﻂ ﻓﺘﻮﮔﺮاﻣﺘﺮی‬ ‫ﺟﻬﺖ ارﺗﺒﺎط ﺑﻴﻦ ﺗﺼﻮﻳﺮ و زﻣﻴﻦ‬

‫‪Analytical Photogrammetry‬‬

‫اراﺋﻪ دهﻨﺪﻩ ‪:‬‬

‫ﺻﻔﺎ ﺧﺰاﺋﻲ‬ ‫‪E-mail: [email protected]‬‬ ‫ﻧﻴﻤﺴﺎل ﺗﺤﺼﻴﻠﯽ ‪٨۵-٨۶-٢ :‬‬ ‫‪1‬‬

‫‪2‬‬

‫ﻣﻌﺎدﻻت هﻢ ﺧﻄﯽ‬

‫ﮐﻤﭙﺎراﺗﻮر‬ ‫ﺗﻮﺟﻴﻪ داﺧﻠﯽ‬

‫‪Collinearity Equations‬‬

‫ﺑﺎ ﻓﺮض ﺡﺬف اﻋﻮﺟﺎﺟﺎت و اﻧﺠﺎم ﺗﻮﺟﻴﻪ داﺧﻠﻲ‬ ‫ﺗﻮﺟﻴﻪ ﻧﺴﺒﯽ‬

‫ﺗﺼﻮﻳﺮ‬

‫ﺗﺒﺪﻳﻼت ‪ 2D‬و ‪3D‬‬

‫ﻣﺪل‬

‫ﺗﻮﺟﻴﻪ ﻣﻄﻠﻖ‬

‫‪4‬‬

‫ﻣﻌﺎدﻻت رﻳﺎﺿﯽ ﺗﺤﺖ‬ ‫ﺷﺮاﻳﻂ ﻓﺘﻮﮔﺮاﻣﺘﺮی‬

‫زﻣﻴﻦ‬

‫‪3‬‬

‫)‪mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007‬‬

‫‪1‬‬

z

OA = k* R * Oa (kϕw) Oa

y

κ

= λ * M * OA ( wϕk )

φ

O (X0 Y0 Z0)

-c

ω x

y x

Z a (x, y, –c)

Y A (X Y Z) ٥

X ٦

 x a   x p   xa − x p  oa =  ya  −  y p  =  ya − y p  − c   0   − c  →

Oa

OA

٧

٨

mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007)

2

Oa

= λ * M * OA ( wϕk )

.‫ ﻣﻌﺎدﻝﻪ اول را ﺑﺮ ﻣﻌﺎدﻝﻪ ﺱﻮم ﺗﻘﺴﻴﻢ ﻣﻲ آﻨﻴﻢ‬٢ ‫ﺡﺎل‬ ٩

١٠

‫ﻣﻌﺎدﻻت هﻢ ﺧﻄﯽ‬ Collinearity Condition

‫ﺵﺮط هﻢ ﺧﻄﯽ‬

 m ( X − X 0 ) + m12 (Y − Y0 ) + m13 (Z − Z0 )  x = xo − f  11   m31( X − X 0 ) + m32 (Y − Y0 ) + m33 (Z − Z0 )  m ( X − X 0 ) + m22 (Y − Y0 ) + m23 (Z − Z0 )  y = yo − f  21   m31( X − X 0 ) + m32 (Y − Y0 ) + m33 (Z − Z0 ) 

‫ﻣﻌﺎدﻻت ﻣﻌﻜﻮس‬

Z0

 m (x − xo ) + m21( y − yo ) + m31(− f )  X = X0 + (Z − Z0 ) ⋅  11   m13(x − xo ) + m23( y − yo ) + m33(− f )

‫ ﺗﺎ‬۶ ‫ ﺗﺎ‬٣

‫ﺗﻌﺪاد پﺎراﻣﺘﺮهﺎ ؟‬

 m (x − xo ) + m22( y − yo ) + m32(− f ) Y = Y0 + (Z − Z0 ) ⋅  12   m13(x − xo ) + m23( y − yo ) + m33(− f ) 

‫ﺡﺪاﻗﻞ ﺗﻌﺪاد ﻧﻘﻄﻪ آﻨﺘﺮل ﻣﻮرد ﻧﻴﺎز ؟‬

X0 Y0 ١٢

11

mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007)

3

DLT ‫ﺗﻔﺎوت ﺗﻌﺪاد پﺎراﻣﺘﺮهﺎی ﺷﺮط هﻢ ﺧﻄﯽ و‬ ‫ﺟﻬﺖ ارﺗﺒﺎط ﺑﻴﻦ ﺗﺼﻮﻳﺮ و زﻣﻴﻦ‬

‫ﻣﻌﺎدﻻت ﺵﺮط هﻢ ﺧﻄﻲ‬

x − x0 = − f

M1 X M3X

M X y − y0 = − f 2 M3X

x =

 m11 M = m21 m31

m12 m22 m32

m13  M 1 m23  M 2 m33  M 3

.‫ ﭘﺎراﻣﺘﺮ اﺱﺖ‬٩ ‫ ﭘﺎراﻣﺘﺮ وﻝﻲ ﻣﻌﺪﻻت ﺵﺮط هﻢ ﺧﻄﻲ داراي‬١١ ‫ دارای‬DLT ‫ درﺣﺎﻝﻴﻜﻪ‬،‫ ﻓﻀﺎی ﺧﺎم دو ﺑﻌﺪی)آﻤﭙﺎراﺗﻮر( را ﺑﻪ ﻓﻀﺎی ﺱﻪ ﺑﻌﺪی)زﻣﻴﻦ( ﻣﺮﺗﺒﻂ ﻣﯽ ﻧﻤﺎیﺪ‬DLT ‫ ﺑﻨﺎﺑﺮایﻦ اﺧﺘﻼف در‬.‫ ﻓﻀﺎی ﺗﺼﻮیﺮ را ﺑﻪ ﻓﻀﺎی ﺱﻪ ﺑﻌﺪی زﻣﻴﻦ ارﺗﺒﺎط ﻣﻲ دهﺪ‬،‫ﺵﺮط هﻢ ﺧﻄﯽ‬ .‫ﻓﻀﺎی ﺗﺼﻮیﺮ اﺱﺖ‬ .‫ ﻣﯽ ﺑﺎﺵﺪ‬nonnon-orthogonality ‫ و‬affinity ‫ ﭘﺎراﻣﺘﺮ‬٢ ‫ﺑﻨﺎﺑﺮایﻦ اﺧﺘﻼف در‬

X − X0 X =  Y − Y0   Z − Z 0 

a1 X + a 2Y + a 3 Z + a 4 c1 X + c 2Y + c 3 Z + 1

b X + b 2Y + b3 Z + b 4 y = 1 c1 X + c 2 Y + c 3 Z + 1

δα

„

„

λx ≠ λ y

‫ ﺑﺮای دورﺑﻴﻨﻬﺎی ﻏﻴﺮ ﻣﺘﺮیﮏ‬-‫ ﺱﺮیﻊ و ﺁﺱﺎن ﺑﻮدﻩ – هﻤﻴﺸﻪ ﺟﻮاﺑﮕﻮ ﻧﻴﺴﺖ‬-‫ ﺧﻄﯽ اﺱﺖ‬:DLT .‫ دﻗﻴﻖ ﺑﻮدﻩ و هﻤﻮارﻩ ﺟﻮاﺑﮕﻮ هﺴﺘﻨﺪ‬-‫ ﻏﻴﺮ ﺧﻄﯽ اﻧﺪ و ﻧﻴﺎز ﺑﻪ ﺧﻄﯽ ﺱﺎزی دارﻧﺪ‬: ‫ﺵﺮط هﻢ ﺧﻄﯽ‬

DLT Transformation

„

„ „

١٣ 14

Collinearity Equation Linearization

‫ﺧﻄﻲ ﺱﺎزي ﻣﻌﺎدﻻت هﻢ ﺧﻄﻲ‬ ‫ﺑﻪ آﻤﻚ ﺑﺴﻂ ﺱﺮي ﺗﻴﻠﻮر‬

‫ﻣﻌﺎدﻝﻪ اول‬

Tayler power series very good approximation

 m ( X − X 0 ) + m12 (Y − Y0 ) + m13 (Z − Z0 )  x = x p − f  11   m31( X − X 0 ) + m32 (Y − Y0 ) + m33 (Z − Z0 )

c11 =  m ( X − X 0 ) + m22 (Y − Y0 ) + m23 (Z − Z0 ) y = y p − f  21   m31( X − X 0 ) + m32 (Y − Y0 ) + m33 (Z − Z0 ) 

f [r ⋅ (cos φ ⋅ ∆X + sin ω ⋅ sin φ ⋅ ∆Y − cos ω ⋅ sin φ ⋅ ∆Z ) q2 − q ⋅ (− sin φ ⋅ cos κ ⋅ ∆X + sin ω ⋅ cos φ ⋅ cos κ ⋅ ∆Y − cos ω ⋅ cos φ ⋅ cos κ ⋅ ∆Z )]

c12 =

‫ ؟‬f ‫ﻣﺸﺘﻖ ﻧﺴﺒﺖ ﺑﻪ‬

dx = c11dω + c12 dφ + c13 dκ − c14 dX 0 − c15 dY0 − c16 dZ 0 + c14 dX + c15 dY + c16 dZ dy = c21dω + c22 dφ + c23 dκ − c24 dX 0 − c25 dY0 − c26 dZ 0 + c24 dX + c25 dY + c26 dZ dX0, dY0, dZ0 = corrections to exposure station location dXA, dYA, dZA = corrections to ground coordinates

r q

f [r ⋅ (− m 33 ⋅ ∆Y + m 32 ⋅ ∆Z ) − q ⋅ ( −m13 ⋅ ∆Y + m12 ⋅ ∆Z )] q2

c13 = −

where: c’s = coefficients equal to partial derivatives dω, dφ, dκ = corrections to angular attitude

x − xo = − f

f (m 21 ⋅ ∆X + m 22 ⋅ ∆Y + m 23 ⋅ ∆Z ) q

f (r ⋅ m31 − q ⋅ m11) q2 f c15 = 2 (r ⋅ m 32 − q ⋅ m12) q f c16 = 2 (r ⋅ m33 − q ⋅ m13) q c14 =

١٥

١٦

mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007)

4

‫‪s‬‬ ‫‪q‬‬

‫ﺡﻞ ﻣﺴﺎﺋﻞ ﻓﺘﻮﮔﺮاﻣﺘﺮی‬

‫‪y − yo = − f‬‬

‫ﻣﻌﺎدﻝﻪ دوم‬

‫‪f‬‬ ‫]) ‪c21 = 2 [ s ⋅ ( − m33 ⋅ ∆Y + m32 ⋅ ∆Z ) − q ⋅ ( − m 23 ⋅ ∆Y + m 22 ⋅ ∆Z‬‬ ‫‪q‬‬

‫„ ﻣﺴﺎﺋﻞ ﻣﺮﺑﻮط ﺑﻪ یﮏ ﻋﮑﺲ‬ ‫„ ﻣﺴﺎﺋﻞ ﻣﺮﺑﻮط ﺑﻪ یﮏ زوج ﻋﮑﺲ‬ ‫„ ﻣﺴﺎﺋﻞ ﻣﺮﺑﻮط ﺑﻪ ﺑﻴﺶ از یﮏ زوج ﻋﮑﺲ )ﻣﺜﻠﺚ ﺑﻨﺪی(‬

‫‪f‬‬ ‫) ‪[ s ⋅ (cos φ ⋅ ∆X + sin ω ⋅ sin φ ⋅ ∆Y − cos ω ⋅ sin φ ⋅ ∆Z‬‬ ‫‪q2‬‬ ‫]) ‪− q ⋅ (− sin φ ⋅ sin κ ⋅ ∆X − sin ω ⋅ cos φ ⋅ sin κ ⋅ ∆Y + cos ω ⋅ cos φ ⋅ sin κ ⋅ ∆Z‬‬

‫= ‪c22‬‬

‫‪f‬‬ ‫) ‪( m11 ⋅ ∆X + m12 ⋅ ∆Y + m13 ⋅ ∆Z‬‬ ‫‪q‬‬ ‫‪f‬‬ ‫)‪c24 = 2 ( s ⋅ m31 − q ⋅ m 21‬‬ ‫‪q‬‬ ‫‪f‬‬ ‫)‪c25 = 2 ( s ⋅ m32 − q ⋅ m 22‬‬ ‫‪q‬‬ ‫= ‪c23‬‬

‫‪f‬‬ ‫)‪( s ⋅ m33 − q ⋅ m 23‬‬ ‫‪q2‬‬

‫= ‪c26‬‬

‫‪١٧‬‬ ‫‪18‬‬

‫ﺗﺮﻣﻴﻢ ﺗﺤﻠﻴﻠﻲ ﺑﺎ ﻳﻚ ﻋﻜﺲ‬

‫ﺡﻞ ﻣﺴﺎﺋﻞ ﻓﺘﻮﮔﺮاﻣﺘﺮی‬ ‫„ ﻣﺴﺎﺋﻞ ﻣﺮﺑﻮط ﺑﻪ یﮏ ﻋﮑﺲ‬

‫ﺗﺮﻣﻴﻢ ﺗﺤﻠﻴﻠﻲ ‪ :‬ﺗﻬﻴﻪ ﻧﻘﺸﻪ از ﻳﮏ ﻋﮑﺲ‬

‫در اﺑﺘﺪا ﺑﺎیﺴﺘﯽ دو کﺎر ذیﻞ ﺗﻮاﻣﺎ اﻧﺠﺎم ﮔﺮدﻧﺪ ‪:‬‬

‫ ﺣﺨﻒ ﺧﻄﺎهﺎی ﺱﻴﺴﺘﻤﺎﺗﻴﮏ)اﻋﻮﺟﺎﺟﺎت(‬‫‪ -‬ﺗﻮﺟﻴﻪ داﺧﻠﯽ‪ :‬از ﺱﻴﺴﺘﻢ ﻣﺨﺘﺼﺎت کﻤﭙﺎراﺗﻮر ﺑﻪ ﻋﮑﺲ ﺑﺮویﻢ‬

‫در ﻣﺮﺡﻠﻪ ﺑﻌﺪ ‪:‬‬ ‫‪ -١‬ﻣﺤﺎﺱﺒﻪ اﻝﻤﺎﻧﻬﺎی ﺗﻮﺟﻴﻪ ﺧﺎرﺟﯽ دورﺑﻴﻦ )‪(EOP‬‬ ‫‪ -٢‬ﻣﺤﺎﺱﺒﻪ ﻣﺨﺘﺼﺎت زﻣﻴﻨﯽ هﺮ ﻧﻘﻄﻪ اﺧﺘﻴﺎری‬

‫‪20‬‬

‫ﺗﺮﻓﻴﻊ ﻓﻀﺎﻳﯽ‬

‫ﺗﺮﻣﻴﻢ ﺗﺤﻠﻴﻠﻲ‬

‫ﺗﻘﺎﻃﻊ ﻓﻀﺎﻳﯽ‬

‫‪19‬‬

‫)‪mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007‬‬

‫‪5‬‬

Space Resection

‫ ﺗﺮﻓﻴﻊ ﻓﻀﺎیﯽ‬-١

‫ﻣﺨﺘﺼﺎت ﻧﻘﺎط آﻨﺘﺮل زﻣﻴﻨﻲ ﻣﻌﻠﻮم ﻣﻲ ﺑﺎﺵﺪ‬

0

‫ﻣﻘﺎدﻳﺮ اوﻝﻴﻪ پﺎراﻣﺘﺮهﺎی ﻣﺠﻬﻮل ؟‬ 0

0

dx = c11dω + c12 dφ + c13 dκ − c14 dX 0 − c15 dY0 − c16 dZ 0 + c14 dX + c15 dY + c16 dZ 0

0

dy = c21dω + c22 dφ + c23 dκ − c24 dX 0 − c25 dY0 − c26 dZ 0 + c24 dX + c25 dY + c26 dZ ١ ‫ﻧﻘﻄﻪ‬

 dx1   c11 c12  dy  c  1   21 c22  .   . .  = . . .    dxn   c11 c12    dyn  c21 c22

c13 c23

− c14 − c24

− c15 − c25

.

.

.

.

.

.

c13 c23

− c14 − c24

− c15 − c25

2n × 1

− c16   dω  − c26   dϕ  .   dk  .  .  dX 0    − c16 dY0    − c26   dZ 0  2n × 6

n ‫ﻧﻘﻄﻪ‬

0

ω0 = ϕ 0 = 0 2D similarity ‫ﺗﺒﺪﻳﻞ‬

a     X   x y 1 0  b   Y  =  y − x 0 1  c        d 

X = ( A T . A) −1 . AT .L

λ0 = a 2 + b 2 b K 0 = Arc tan( ) a

X0 = c Y0 = d K 0 = AZ AB − AZ ab = Arc tan(

YB − Y A y − yA ) − Arc tan( B ) XB − XA xB − x A

−

Z 0 = λ. f + h AB

6× 1

−

‫ ﺗﺎ‬٣ ‫ﺡﺪاﻗﻞ ﻧﻘﺎط آﻨﺘﺮل ﻣﻮرد ﻧﻴﺎز ؟‬

Z 0 = H − h AB

∧

δ X = ( A T . A) −1 A.T δL

٢١

٢٢

‫ ﻣﺤﺎﺱﺒﻪ ﻣﺨﺘﺼﺎت زﻣﻴﻨﯽ هﺮ ﻧﻘﻄﻪ اﺧﺘﻴﺎری‬-٢

‫ ﻣﺤﺎﺱﺒﻪ ﻣﺨﺘﺼﺎت زﻣﻴﻨﯽ هﺮ ﻧﻘﻄﻪ اﺧﺘﻴﺎری‬-٢

‫ﻣﻌﺎدﻻت ﻣﻌﻜﻮس هﻢ ﺧﻄﻲ‬

‫ﻣﻮﻗﻌﻴﺖ و وﺿﻌﻴﺖ ﻣﺮآﺰ ﺗﺼﻮﻳﺮ ﻣﻌﻠﻮم اﺱﺖ‬ O a

 m (x − xo ) + m21( y − yo ) + m31(− f )  X = X0 + (Z − Z0 ) ⋅  11   m13(x − xo ) + m23( y − yo ) + m33(− f )  m (x − xo ) + m22( y − yo ) + m32(− f ) Y = Y0 + (Z − Z0 ) ⋅  12   m13(x − xo ) + m23( y − yo ) + m33(− f )  ‫( ﻣﻮﺟﻮد ﺑﺎﺷﺪ‬DTM) ‫ ﺑﺎﻳﺪ ﻣﺪل رﻗﻮﻣﻲ ارﺗﻔﺎﻋﻲ زﻣﻴﻦ‬: ‫ﺑﺮاي ﺡﻞ‬

Terrain Surface A2

‫زﻣﻴﻦ ﻣﺴﻄﺢ ﺑﺎﺷﺪ‬

‫ﺗﻘﺎﻃﻊ‬ A

Vertical View Plane

X=

a1 x + a2 y + a3 c1 x + c2 y + 1

Y=

b1 x + b2 y + b3 c1 x + c2 y + 1

‫ اﺱﺎس ﺗﺮﻣﻴﻢ ﺗﺤﻠﻴﻠﯽ‬2D Projective

: ‫ﺑﻨﺎﺑﺮایﻦ‬ ‫ ﻣﻌﺎدﻻت ﭘﺮوژآﺘﻴﻮ دو ﺑﻌﺪي ) اﺣﺘﻴﺎﺟﻲ ﺑﻪ ﺗﻮﺟﻴﻪ داﺧﻠﻲ ﻧﺪاﺵﺘﻪ و ﻣﺴﺘﻘﻴﻤﺎ از‬: ‫• اﮔﺮ زﻣﻴﻦ ﻣﺴﻄﺢ ﺑﺎﺵﺪ‬ (‫ﻣﺨﺘﺼﺎت دﺱﺘﮕﺎهﻲ ﺑﻪ زﻣﻴﻦ ﻣﻲ رﺱﻴﻢ‬ A1

Z0 ٢٣

‫ ﻣﻌﺎدﻻت ﺵﺮط هﻢ ﺧﻄﻲ‬: ‫• اﮔﺮ زﻣﻴﻦ ﻧﺎ ﻣﺴﻄﺢ ﺑﺎﺵﺪ‬

٢٤

mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007)

6

‫ﺗﺮﻣﻴﻢ ﺗﺤﻠﻴﻠﯽ ﺑﺎ اﺱﺘﻔﺎدﻩ از ﻳﮏ زوج ﻋﮑﺲ‬

‫ﺗﺮﻣﻴﻢ ﺗﺤﻠﻴﻠﻲ ﺑﺎ ﻳﻚ زوج ﻋﻜﺲ‬

‫ﺗﺮﻓﻴﻊ ﻓﻀﺎﻳﯽ‬

(EOP) ‫ ﻣﺤﺎﺱﺒﻪ اﻝﻤﺎﻧﻬﺎی ﺗﻮﺟﻴﻪ ﺧﺎرﺟﯽ دورﺑﻴﻦ‬-١

‫ﺗﻘﺎﻃﻊ ﻓﻀﺎﻳﯽ‬

‫ ﻣﺤﺎﺱﺒﻪ ﻣﺨﺘﺼﺎت زﻣﻴﻨﯽ هﺮ ﻧﻘﻄﻪ اﺧﺘﻴﺎری‬-٢

‫ ﺗﺮﻓﻴﻊ و ﺗﻘﺎﻃﻊ ﻓﻀﺎیﻲ‬: ‫راﻩ ﺣﻞ اول‬ ‫ اﻧﺠﺎم ﻣﺮاﺣﻞ ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ و ﻣﻄﻠﻖ‬: ‫راﻩ ﺣﻞ دوم‬

25

26

‫ ﺗﺮﻓﻴﻊ ﻓﻀﺎیﯽ‬-١

‫ ﺗﻘﺎﻃﻊ ﻓﻀﺎیﯽ‬-٢

‫ﻣﺸﺨﺺ آﺮدن ﻣﻮﻗﻌﻴﺖ و وﺽﻌﻴﺖ ﻣﺮاآﺰ ﺗﺼﻮیﺮ دو ﻋﻜﺲ‬ c c  dx'1    dy '  c c  1  .  .   .   .  .  .   ' '  dx'n   c11 c12  c ' c '   dy 'n  =  21 22 0  dx"   0  1  0 0  dx"1   0  .  0    0  .  0 dx"n   0 0 dy"n   0 0 ' 11 ' 21

' 12 ' 22

c c

−c −c

−c −c

−c −c

. .

. .

. .

. .

c13' ' c23

− c14' ' − c24

− c15' ' − c25

0 0 0

0 0 0

0 0 0

0 0 0

0 0

0 0

0 0

0 0

c11"

c12"

c13"

0

" 21

" 22

" 23

' 13 ' 23

0

' 14 ' 24

0

' 15 ' 25

0

' 16 ' 26

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 − c16' ' − c26 (n) 0

0 0

0 0

0 0

0 0

c11" c12" " " c21 c22

c13" " c23

− c14" " − c24

− c15" " − c25

(1)

c

c

c

− c14"

− c15"

−c

−c

" 24

4n×1 ‫ ﺗﺎ‬٣ ‫ﺡﺪاﻗﻞ ﻧﻘﺎط آﻨﺘﺮل ﻣﻮرد ﻧﻴﺎز ؟‬

" 25

0

0    dω '  0  dϕ '  0    dk '  0    dX '0  0   dY '0  0   dZ '  0 .  − c16"   dω"   "  − c26  dϕ "    dk "  (1)     dX "0  (n)   "  dY "0 − c16     dZ "0  "  − c26 

0

0

0

∧

0

0

0

0

0

0

0

dy = c21dω + c22 dφ + c23 dκ − c24 dX 0 − c25 dY0 − c26 dZ 0 + c24 dX + c25 dY + c26 dZ

 dx '   c14'  dy '   '    c 24  .   .  =  .   .  dx "   c14"    "  dy "   c 24

c16'  '  c 26   dX .   . dY .   dZ "   c16  " c 26 

c15' ' c 25 .

. c15" " c 25

    3×1

4n×3

4n×1 ∧

4n×12

δ X = ( A T . A) −1 A.T δL

0

dx = c11dω + c12 dφ + c13 dκ − c14 dX 0 − c15 dY0 − c16 dZ 0 + c14 dX + c15 dY + c16 dZ

δ X = ( A . A) A.T δL ٢٧

T

−1

٢٨

mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007)

7

‫ﻣﻘﺎدﻳﺮ اوﻝﻴﻪ پﺎراﻣﺘﺮهﺎی ﻣﺠﻬﻮل ؟‬

‫راﻩ ﺣﻞ دوم ‪:‬‬ ‫اﻧﺠﺎم ﻣﺮاﺣﻞ ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ و ﻣﻄﻠﻖ ﺑﻄﻮر ﺟﺪاﮔﺎﻧﻪ‬

‫‪B = ( X "0 − X '0 ) 2 + (Y "0 −Y '0 ) 2‬‬ ‫"‪p = x'− x‬‬

‫‪O2‬‬

‫'‪B.x‬‬ ‫‪P‬‬ ‫' ‪x” Y = B. y‬‬ ‫‪P‬‬ ‫=‪X‬‬

‫ﺗﻮﺟﻴﻪ ﻧﺴﺒﯽ ﺑﺎ ﺷﺮط هﻢ ﺧﻄﯽ‬

‫‪B‬‬

‫”‪y‬‬

‫’‪x‬‬ ‫”‪a‬‬

‫‪O1‬‬

‫’‪y‬‬ ‫’‪a‬‬

‫‪B. f‬‬ ‫‪Z = Z '0 −‬‬ ‫‪p‬‬

‫‪A‬‬

‫‪29‬‬ ‫‪30‬‬

‫ﺗﻮﺟﻴﻪ ﻧﺴﺒﯽ ﺑﺎ ﺷﺮط هﻢ ﺧﻄﯽ‬

‫‪b‬‬

‫‪O2‬‬

‫در ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ‪ ،‬هﺪف ‪ :‬ﻣﺸﺨﺺ آﺮدن وﺽﻌﻴﺖ ﻧﺴﺒﻲ دو دورﺑﻴﻦ‬

‫’‪y‬‬

‫”‪y‬‬

‫در ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ اﺑﺘﺪا ﺑﺎیﺴﺘﻲ ﻣﺸﺨﺺ آﻨﻴﻢ آﻪ از ﭼﻪ روﺵﻲ اﺱﺘﻔﺎدﻩ ﻣﻲ آﻨﻴﻢ ‪ :‬یﻜﻄﺮﻓﻪ یﺎ دو ﻃﺮﻓﻪ‬ ‫ﻓﺮض ‪ :‬یﻜﻄﺮﻓﻪ‬ ‫ﺑﻨﺎﺑﺮایﻦ ‪ ۶ :‬اﻝﻤﺎن ﺗﻮﺟﻴﻪ ﺧﺎرﺟﻲ را ﺑﺮاي ﻋﻜﺲ ﺱﻤﺖ راﺱﺖ یﺎ ﭼﭗ را ﺙﺎﺑﺖ ﻓﺮض ﻣﻴﻜﻨﻴﻢ‪ .‬ﺑﻨﺎﺑﺮایﻦ‬ ‫ﻣﺸﺘﻘﺎت ﺟﺰﺋﻲ ﺁﻧﻬﺎ ﺻﻔﺮ اﺱﺖ‬

‫”‪x‬‬

‫‪O1‬‬

‫’‪x‬‬ ‫”‪a‬‬

‫’‪a‬‬

‫‪dx' = c'14 dX + c'15 dY + c'16 dZ‬‬

‫ﺑﺮاي ﻋﻜﺲ ﺱﻤﺖ ﭼﭗ‬ ‫ﺑﺮاي ﻋﻜﺲ ﺱﻤﺖ راﺱﺖ‬

‫‪dy ' = c '24 dX + c'25 dY + c'26 dZ‬‬

‫‪0‬‬

‫" ‪dx" = c"11 dω"+c"12 dφ "+c"13 dκ "−c"14 dX "0 −c"15 dY "0 −c"16 dZ "0 +c"14 dX "+c"15 dY "+c"16 dZ‬‬

‫‪0‬‬

‫" ‪dy" = c"21 dω"+c"22 dφ "+c"23 dκ "−c"24 dX "0 −c"25 dY "0 −c"26 dZ "0 +c"24 dX "+c"25 dY "+c"26 dZ‬‬

‫‪A‬‬ ‫ﺗﻌﺪاد ﻣﻌﺎدﻻت ‪4n :‬‬ ‫ﺗﻌﺪاد ﻣﺠﻬﻮﻻت ‪3n+5 :‬‬

‫ﺣﺪاﻗﻞ ﺗﻌﺪاد ﻧﻘﻄﻪ آﻨﺘﺮل ﻣﻮرد ﻧﻴﺎز ‪ ۵ :‬ﺗﺎ‬ ‫‪٣١‬‬

‫‪32‬‬

‫)‪mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007‬‬

‫‪8‬‬

‫ﻣﻘﺎدﻳﺮ اوﻝﻴﻪ پﺎراﻣﺘﺮهﺎی ﻣﺠﻬﻮل ؟‬ ‫در ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ ﺑﺎ ﺵﺮط هﻢ ﺧﻄﻲ ﭼﻨﺎﻧﭽﻪ از ‪ ٢٠‬ﻧﻘﻄﻪ اﺱﺘﻔﺎدﻩ آﻨﻴﻢ‪ ،‬ﺗﻌﺪاد ﻣﺠﻬﻮﻻت ؟‬ ‫اﺑﻌﺎد ﻣﺎﺗﺮیﺲ ﻧﺮﻣﺎل ؟‬

‫‪3*20+5=65‬‬

‫‪w" = 0‬‬

‫‪ϕ"= 0‬‬

‫)‪(65)* (65‬‬ ‫' ‪a 'b‬‬

‫‪− AZ‬‬

‫" ‪a "b‬‬

‫‪O2‬‬

‫‪k " = AZ‬‬

‫‪b m = arbitary‬‬ ‫‪= bm‬‬

‫ﻣﻌﺎیﺐ ایﻦ روش ؟‬ ‫‪ -١‬اﺑﻌﺎد ﺑﺰرگ ﻣﺎﺗﺮیﺲ ﻧﺮﻣﺎل‬ ‫‪ -٢‬ﺑﺪﺱﺖ ﺁوردن ﻣﻘﺎدیﺮ ﺗﻘﺮیﺒﻲ)اوﻝﻴﻪ( ﻧﻴﺎز اﺱﺖ‬

‫?‪H‬‬

‫”‪y‬‬ ‫”‪x‬‬

‫’‪x‬‬ ‫”‪a‬‬

‫’‪y‬‬ ‫’‪a‬‬

‫‪Z‬‬

‫‪A‬‬

‫‪33‬‬

‫در ﺵﺮط هﻢ ﺧﻄﻲ ﻓﺮض ﻣﺎ ﺑﺮ ایﻦ ﺑﻮد آﻪ ﭘﺎﻻیﺶ ﻋﻜﺴﻲ و ﺗﻮﺟﻴﻪ داﺧﻠﻲ اﻧﺠﺎم ﮔﺮﻓﺘﻪ اﺱﺖ‪.‬‬ ‫اﻣﺎ ﭼﻨﺎﻧﭽﻪ از دورﺑﻴﻨﻬﺎي ﻏﻴﺮ ﻣﺘﺮیﻚ دیﺠﻴﺘﺎﻝﻲ)در ﻓﺘﻮﮔﺮاﻣﺘﺮي ﺑﺮد آﻮﺗﺎﻩ( اﺱﺘﻔﺎدﻩ ﻧﻤﺎﺋﻴﻢ‪ ،‬ﺑﺎیﺴﺘﻲ‬ ‫ﭘﺎراﻣﺘﺮهﺎي ﺗﻮﺟﻴﻪ داﺧﻠﻲ و ﻧﻴﺰ اﻋﻮﺟﺎﺟﺎت ﺵﻌﺎﻋﻲ و ﻣﻤﺎﺱﻲ ﻝﻨﺰ را ﻧﻴﺰ ﺑﻪ ﻋﻨﻮان ﻣﺠﻬﻮﻻت در ﻧﻈﺮ‬ ‫ﺑﮕﻴﺮیﻢ‪.‬‬ ‫ﺵﺮط هﻢ ﺧﻄﻲ ﺑﺎ ﭘﺎراﻣﺘﺮهﺎي اﺽﺎﻓﻲ‬

‫آﺎﻝﻴﺒﺮاﺱﻴﻮن دورﺑﻴﻨﻬﺎي ﻣﺘﺮیﻚ ‪:‬‬ ‫‪ -١‬ﻓﺎﺻﻠﻪ آﺎﻧﻮﻧﻲ )اﺻﻠﻲ(‬ ‫‪ -٢‬ﻣﻮﻗﻌﻴﺖ ﻧﻘﻄﻪ اﺻﻠﻲ ﻧﺴﺒﺖ ﺑﻪ ﻓﻴﺪوﺵﺎل ﻣﺎرآﻬﺎ‬ ‫‪ -٣‬ﻣﺨﺘﺼﺎت ﻓﻴﺪوﺵﺎل ﻣﺎرآﻬﺎ‬ ‫‪ -۴‬ﻣﺸﺨﺼﺎت اﻋﻮﺟﺎﺟﺎت)ﺵﻌﺎﻋﻲ و ﻣﻤﺎﺱﻲ( ﻝﻨﺰهﺎ‬

‫ﺗﻌﺪاد ﭘﺎراﻣﺘﺮهﺎي ﻣﺠﻬﻮل ؟ ‪6+3+6 = 15‬‬

‫‪36‬‬

‫‪= H‬‬

‫"‬ ‫‪0‬‬

‫" ‪p = x '− x‬‬ ‫' ‪b .x‬‬ ‫‪X = m‬‬ ‫‪P‬‬ ‫'‪b .y‬‬ ‫‪Y = m‬‬ ‫‪P‬‬ ‫‪b .f‬‬ ‫‪Z = H − m‬‬ ‫‪p‬‬

‫ﻣﺰیﺖ ایﻦ روش ؟‬ ‫ﻣﺨﺘﺼﺎت ﻣﺪل هﻤﺰﻣﺎن ﺑﺎ ﺣﻞ ﻣﺠﻬﻮﻻت ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ ﻣﺤﺎﺱﺒﻪ ﻣﻲ ﮔﺮدﻧﺪ‪.‬‬

‫‪34‬‬

‫"‬ ‫‪0‬‬

‫‪X‬‬

‫‪Y 0" = 0 ,...‬‬

‫‪bm‬‬

‫‪O1‬‬

‫‪M1 X‬‬ ‫‪M3X‬‬

‫ﻣﻘﺪﻣﻪ اي ﺑﺮ‬ ‫‪Self Calibration‬‬

‫‪x − x0 + δx = −c‬‬

‫‪M X‬‬ ‫‪y − y0 + δy = −c 2‬‬ ‫‪M3X‬‬ ‫) ‪δx = F (c, x0 , y0 , k1 , k 2 , k3 , p1 , p2 , p3‬‬ ‫) ‪δy = G (c, x0 , y0 , k1 , k 2 , k3 , p1 , p2 , p3‬‬

‫‪35‬‬

‫)‪mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007‬‬

‫‪9‬‬

‫ﺷﺮط هﻢ ﺹﻔﺤﻪ ای‬

‫‪ D.L.T‬ﺑﺎ ﭘﺎراﻣﺘﺮهﺎي اﺽﺎﻓﻲ‬

‫‪Collinearity Condition‬‬

‫هﻨﺪﺱﻪ ‪Epipolar‬‬ ‫„‬ ‫„‬ ‫„‬

‫‪Epipolar plane‬‬ ‫‪Epipolar lines‬‬ ‫‪Conjugated points‬‬

‫‪a1 X + a 2Y + a 3 Z + a 4‬‬ ‫‪c1 X + c 2Y + c 3 Z + 1‬‬

‫= ‪x + δx‬‬

‫‪b1 X + b 2 Y + b 3 Z + b 4‬‬ ‫‪c1 X + c 2 Y + c 3 Z + 1‬‬

‫= ‪y + δy‬‬

‫) ‪δx = F (k1 , k 2 , k3 , p1 , p2 , p3‬‬ ‫) ‪δy = G (k1 , k 2 , k3 , p1 , p2 , p3‬‬

‫ﺻﻔﺤﻪ ﮔﺬرﻧﺪﻩ از ﻣﺮاآﺰ ﺗﺼﻮیﺮ و ﻧﻘﺎط‬ ‫ﻧﻈﻴﺮ ﻋﻜﺴﻲ در یﻚ ‪stereopair‬‬ ‫ﻣﺤﻞ ﺗﻼﻗﻲ ﺻﻔﺤﻪ اﭘﻲ ﭘﻮﻻر ﺑﺎ ﻋﻜﺲ هﺎ‬

‫ﺗﻌﺪاد ﭘﺎراﻣﺘﺮهﺎي ﻣﺠﻬﻮل ؟ ‪11+6 = 17‬‬

‫ﺱﻴﺴﺘﻢ ﻣﺨﺘﺼﺎت ﻣﺪل‬ ‫‪37‬‬

‫‪38‬‬

‫ﻣﻌﺎدﻻت هﻢ ﺻﻔﺤﻪ اي‬ ‫راﻩ ﺣﻞ اول ‪:‬‬

‫ﺷﺮط هﻢ ﺹﻔﺤﻪ ای‬

‫‪aX 0' + bY0' + cZ 0' + d = 0‬‬ ‫‪aX 0" + bY0" + cZ 0" + d = 0‬‬

‫‪Y‬‬

‫‪ax + by + cz + d = 0‬‬ ‫'‬

‫'‬

‫ﭼﻨﺪ ﺻﻔﺤﻪ اﭘﻲ ﭘﻮﻻر داریﻢ ؟‬

‫'‬

‫‪ax " + by " + cz " + d = 0‬‬

‫‪b‬‬

‫‪pc2‬‬

‫‪X‬‬

‫ﺑﺮاي ﻧﺸﺎن دادن یﻚ ﺻﻔﺤﻪ ‪ ٣ ،‬ﻧﻘﻄﻪ آﺎﻓﻴﺴﺖ‪ .‬ﭘﺲ یﻜﻲ‬ ‫از ﻣﻌﺎدﻻت اﺽﺎﻓﻲ اﺱﺖ یﺎ ﻣﺴﺘﻘﻞ ﻧﻴﺴﺖ‬

‫”‪y‬‬ ‫”‪x‬‬

‫‪a ( X − X 0' ) + b(Y0" − Y0' ) + c ( Z 0" − Z 0' ) = 0‬‬ ‫"‬ ‫‪0‬‬

‫‪ax ' + by ' + cz ' + d = 0‬‬

‫’‪x‬‬ ‫”‪a‬‬

‫‪pc1‬‬

‫’‪y‬‬

‫ﺑﻴﻨﻬﺎیﺖ‬

‫در ﭼﻪ ﺻﻮرت ﺻﻔﺤﻪ اﭘﻲ ﭘﻮﻻر ایﺠﺎد ﺧﻮاهﺪ ﺵﺪ؟ اﻧﺠﺎم ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ‬ ‫ﭘﺲ ﺵﺮط هﻢ ﺻﻔﺤﻪ اي دﻗﻴﻘﺎ ﻣﺨﻼف ﺵﺮط هﻢ ﺧﻄﻲ اﺱﺖ!‬

‫’‪a‬‬

‫‪ax" + by " + cz " + d = 0‬‬

‫‪Z‬‬

‫ﺑﻲ ﻧﻬﺎیﺖ ﺻﻔﺤﻪ اﭘﻲ ﭘﻮﻻر داریﻢ‪ .‬ﭘﺲ‬ ‫دﺗﺮﻣﻴﻨﺎن ﺽﺮایﺐ ﻣﻌﺎدﻻت ﺑﺎیﺴﺘﻲ ﺻﻔﺮ ﺑﺎﺵﺪ‬ ‫‪bz‬‬ ‫‪b = bx2 + by2 + bz2‬‬

‫‪٤٠‬‬

‫‪by‬‬

‫‪∆Z‬‬

‫‪bx‬‬

‫‪x' y ' z ' = 0‬‬ ‫"‪x" y" z‬‬

‫‪or‬‬

‫‪z' = 0‬‬ ‫"‪z‬‬

‫‪∆Y‬‬

‫‪∆X‬‬

‫'‪y‬‬ ‫"‪y‬‬

‫'‪x‬‬ ‫"‪x‬‬

‫‪A‬‬

‫ﻣﻌﺎدﻻت ﺵﺮط هﻢ ﺻﻔﺤﻪ اي‬ ‫‪39‬‬

‫)‪mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007‬‬

‫‪10‬‬

‫ﻣﻌﺎدﻻت هﻢ ﺻﻔﺤﻪ اي‬

‫راﻩ ﺣﻞ ﺱﻮم ‪:‬‬

‫ﻣﻌﺎدﻻت هﻢ ﺻﻔﺤﻪ اي‬

‫راﻩ ﺣﻞ دوم ‪:‬‬

‫→‬

‫→‬

‫→‬

‫‪b − R1 + R2 = 0‬‬

‫→‬

‫→‬

‫→‬

‫‪pc2‬‬

‫‪b .( R1× R 2 ) = 0‬‬ ‫‪by‬‬

‫‪bx‬‬

‫‪z1 = 0‬‬

‫‪y1‬‬

‫‪x1‬‬

‫‪bz‬‬ ‫‪z2‬‬

‫‪y2‬‬

‫‪x2‬‬

‫‪b‬‬

‫”‪y‬‬ ‫”‪x‬‬

‫ﻣﻌﺎدﻻت ﺵﺮط هﻢ ﺻﻔﺤﻪ اي‬

‫’‪x‬‬ ‫”‪a‬‬

‫’‪a‬‬

‫‪R2‬‬

‫‪R1‬‬

‫ﭘﺎراﻣﺘﺮهﺎي ﻣﺠﻬﻮل )در ﺗﻘﺎﻃﻊ ﻓﻀﺎیﻲ(؟‬

‫→‬

‫‪ x2 ‬‬ ‫‪ x"− x0 ‬‬ ‫→‬ ‫‪b1 =  y2  M 2T  y"− y0 ‬‬ ‫‪ z 2 ‬‬ ‫‪ − f ‬‬

‫→‬

‫‪R2 = λ2 a2‬‬

‫"‬ ‫'‬ ‫‪bx   X 0 − X 0 ‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪b = by  =  Y0" − Y0' ‬‬ ‫"‬ ‫'‬ ‫‪bz   Z 0 − Z 0 ‬‬

‫‪ x1 ‬‬ ‫‪ x'− x0 ‬‬ ‫→‬ ‫‪a1 =  y1  = M 1T  y '− y0 ‬‬ ‫‪ z1 ‬‬ ‫‪ − f ‬‬

‫→‬

‫‪bx ‬‬ ‫‪ x1 ‬‬ ‫‪ x2 ‬‬ ‫‪b  − λ M T  y  + λ M T  y  = 0‬‬ ‫‪ y  1 1  1 2 2  2 ‬‬ ‫‪bz ‬‬ ‫‪ z1 ‬‬ ‫‪ z 2 ‬‬

‫‪A‬‬

‫‪ ١٢‬ﺗﺎ‬ ‫‪٤١‬‬

‫‪bz‬‬

‫ﺧﻄﻲ ﺱﺎزي ﻣﻌﺎدﻻت هﻢ ﺻﻔﺤﻪ اي‬

‫‪bx‬‬

‫‪x' y ' z ' = 0‬‬ ‫"‪x" y" z‬‬

‫‪c5(1)  dbY ‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪c5( 2 )   dbZ ‬‬ ‫‪.   dw ‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪.   dϕ ‬‬ ‫‪( n)  ‬‬ ‫‪‬‬ ‫‪dk‬‬ ‫‪c5  ‬‬ ‫اﺑﻌﺎد ﻣﺎﺗﺮیﺲ ﻧﺮﻣﺎل ؟‬ ‫‪5*5‬‬

‫( ‪F = F0 +‬‬

‫‪c1bY + c2bZ + c3 dw + c4 dϕ + c5 dk + c6 = 0‬‬

‫)‪c4(1‬‬ ‫)‪c4( 2‬‬

‫”‪a‬‬

‫’‪a‬‬

‫‪R2‬‬

‫‪R1‬‬

‫‪bz‬‬ ‫‪z1 = 0‬‬ ‫‪z2‬‬

‫‪by‬‬ ‫‪y1‬‬ ‫‪y2‬‬

‫‪bx‬‬ ‫‪x1‬‬ ‫‪x2‬‬

‫‪A‬‬

‫ﻣﺘﺪ ‪ :‬ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ یﻚ ﻃﺮﻓﻪ‬

‫‪F‬‬

‫)‪c3(1‬‬ ‫) ‪c3( 2‬‬

‫”‪x‬‬

‫’‪x‬‬

‫’‪y‬‬

‫ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ ﺑﺎ ﻣﻌﺎدﻻت هﻢ ﺻﻔﺤﻪ اي‬

‫‪∂F‬‬ ‫‪∂F‬‬ ‫‪∂F‬‬ ‫‪∂F‬‬ ‫‪∂F‬‬ ‫( ‪) 0 dbY +‬‬ ‫"‪) 0 dbZ + ( ) 0 dx"+( ) 0 dy"+( ) 0 dz = 0‬‬ ‫"‪∂z‬‬ ‫"‪∂y‬‬ ‫"‪∂x‬‬ ‫‪∂∆Z‬‬ ‫‪∂∆Y‬‬

‫یﻚ ﻣﻌﺎدﻝﻪ ﺑﺮاي ‪ ٢‬ﻧﻘﻄﻪ ﻧﻈﻴﺮ ﻋﻜﺴﻲ‬

‫→‬

‫”‪y‬‬

‫→‬

‫‪b − λ1 a1 + λ2 a2 = 0‬‬

‫ﻣﻌﺎدﻻت ﺵﺮط هﻢ ﺻﻔﺤﻪ اي‬

‫‪by‬‬

‫‪pc2‬‬

‫‪ ٣‬ﻣﻌﺎدﻝﻪ داریﻢ ﺑﺎ دو ﻣﺠﻬﻮل‬ ‫یﻜﻲ را ﺣﺬف ﻣﻲ آﻨﻴﻢ‪ .‬ﺗﻨﻬﺎ یﻚ ﻣﻌﺎدﻝﻪ ﺑﺎﻗﻲ ﻣﻲ ﻣﺎﻧﺪ‪ .‬ﺁﻧﺮا ﺑﻪ ﺻﻮرت دﺗﺮﻣﻴﻨﺎﻧﻲ ﻣﻲ ﻧﻮیﺴﺴﻢ‬

‫‪٤٢‬‬

‫‪٤٤‬‬

‫‪R1 = λ1 a1‬‬

‫‪b‬‬

‫‪pc1‬‬

‫‪λ1 , λ2‬‬

‫‪ ١٢‬ﭘﺎراﻣﺘﺮﺗﻮﺟﻴﻪ ﺧﺎرﺟﻲ )اﻝﻤﺎﻧﻬﺎي وﺽﻌﻴﺖ و ﻣﻮﻗﻌﻴﺖ دو ﻣﺮآﺰ ﺗﺼﻮیﺮ(‬ ‫ﺣﺪاﻗﻞ ﺗﻌﺪاد ﻧﻘﻄﻪ آﻨﺘﺮل ؟‬

‫→‬

‫‪pc1‬‬

‫’‪y‬‬

‫→‬

‫)‪ c6(1)   c1(1‬‬ ‫)‪ ( 2 )   ( 2‬‬ ‫‪c6  c1‬‬ ‫‪ . = .‬‬ ‫‪  ‬‬ ‫‪ .   .‬‬ ‫) ‪c ( n )  c ( n‬‬ ‫‪ 6  1‬‬

‫)‪c2(1‬‬ ‫) ‪c2( 2‬‬

‫‪.‬‬

‫‪.‬‬

‫‪.‬‬

‫‪.‬‬

‫‪.‬‬

‫‪.‬‬

‫) ‪c4( n‬‬

‫) ‪c3( n‬‬

‫) ‪c2( n‬‬

‫ﻓﺮض ﻣﻲ آﻨﻴﻢ ﺽﺮیﺐ ﻣﻘﻴﺎس دو ﻋﻜﺲ )ﺑﺮاي ﺗﻤﺎم ﻧﻘﺎط دو ﻋﻜﺲ( ﺑﺮاﺑﺮ و یﻜﻲ ﺑﺎﺵﺪ‬ ‫ﻋﻜﺲ ﺱﻤﺖ ﭼﭗ را ﺙﺎﺑﺖ ﻣﻲ آﻨﻴﻢ‬ ‫ﭘﺎراﻣﺘﺮهﺎي ﻣﺠﻬﻮل ؟ ‪ ۵‬ﺗﺎ‬

‫"‪w" , ϕ " , k " , by" , bz‬‬

‫ﺣﺪاﻗﻞ ﺗﻌﺪاد ﻧﻘﺎط آﻨﺘﺮل ﻣﻮرد ﻧﻴﺎز ؟ ‪ ۵‬ﺗﺎ‬

‫ﻣﺮﺣﻠﻪ ﺑﻌﺪ ‪ :‬ﺗﻘﺎﻃﻊ ﻓﻀﺎیﻲ‬

‫∧‬

‫‪δ X = ( A T . A) −1 A.T δL‬‬ ‫‪43‬‬

‫)‪mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007‬‬

‫‪11‬‬

‫ﺷﺮط هﻢ ﺹﻔﺤﻪ اي ﺗﻘﺮﻳﺒﻲ‬

‫ﻣﻘﺎدیﺮ اوﻝﻴﻪ ؟‬

‫ﻓﺮض اول ‪z’=z”=f :‬‬ ‫ﻓﺮض دوم ‪ :‬ﻧﻘﺎط اﺱﺘﺎﻧﺪارد ﻣﺪل )‪ ۶‬ﻧﻘﻄﻪ( ﺑﻪ ﺻﻮرت ﻗﺮیﻨﻪ ﺑﺎﺵﻨﺪ‬ ‫‪4‬‬ ‫ﻓﺮض ﺱﻮم ‪bY = bZ = 0 :‬‬

‫‪3‬‬

‫‪w =ϕ = 0‬‬ ‫' ‪k = AZ a"b" − AZ a 'b‬‬

‫‪2‬‬

‫‪1‬‬

‫‪bY = bZ = 0‬‬

‫‪6‬‬

‫‪5‬‬

‫‪b‬‬ ‫‪d‬‬

‫‪b‬‬ ‫‪ dby ‬‬ ‫‪0 ‬‬ ‫‪dbz ‬‬ ‫‪b ‬‬ ‫‪  dw ‬‬ ‫‪0 ‬‬ ‫‪‬‬ ‫‪dϕ ‬‬ ‫‪b ‬‬ ‫‪  dk ‬‬ ‫‪0‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪bd‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪bd 1 + d 2‬‬

‫‪0‬‬ ‫‪bd‬‬ ‫‪0‬‬

‫‪bd 1 + d 2‬‬ ‫‪bd 1 + d 2‬‬ ‫‪bd 1 + d 2‬‬

‫‪ py1  − b‬‬ ‫‪ py  − b‬‬ ‫‪ 2 ‬‬ ‫‪ py3  − b‬‬ ‫‪=‬‬ ‫‪‬‬ ‫‪ py4  − b‬‬ ‫‪ py5  − b‬‬ ‫‪ ‬‬ ‫‪‬‬ ‫‪ py6  − b‬‬

‫‪1‬‬ ‫]) ‪[( p2 + p4 + p6 ) − ( p1 + p3 + p5‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫]) ‪dw = 2 [2( p1 + p2 ) − ( p3 + p4 + p5 + p6‬‬ ‫‪4d‬‬ ‫‪1‬‬ ‫= ‪dϕ‬‬ ‫]) ‪[( p4 − p6 ) − ( p3 − p5‬‬ ‫‪2bd‬‬ ‫‪1‬‬ ‫= ‪dby‬‬ ‫) ‪dw(3 + 2d 2 ) + ( p2 + p4 + p6‬‬ ‫‪3b‬‬ ‫‪1‬‬ ‫= ‪dbz‬‬ ‫) ‪( p6 − p4‬‬ ‫‪2bd‬‬

‫ﻣﺰیﺖ ﻣﻌﺎدﻻت ﺵﺮط هﻢ ﺧﻄﻲ ﺑﺮ ﺵﺮط هﻢ ﺻﻔﺤﻪ اي در ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ ؟‬ ‫ﭘﺲ از ﺱﺮﺵﻜﻨﻲ ﻣﺪل‪ ،‬ﻣﺪل ﺱﻪ ﺑﻌﺪي ﻣﻌﻠﻮم اﺱﺖ‪ .‬ﭘﺲ اﺣﺘﻴﺎﺟﻲ ﺑﻪ ﺗﻘﺎﻃﻊ ﻓﻀﺎیﻲ ﻧﻴﺴﺖ‪.‬‬

‫= ‪dk‬‬

‫]‬

‫‪46‬‬

‫∧‬

‫ﻣﺰیﺖ ﻣﻌﺎدﻻت ﺵﺮط هﻢ ﺻﻔﺤﻪ اي ﺑﺮ ﺵﺮط هﻢ ﺧﻄﻲ در ﺗﻮﺟﻴﻪ ﻧﺴﺒﻲ ؟‬ ‫ﻧﻴﺎزي ﺑﻪ ﻣﺤﺎﺱﺒﺎت ﭘﻴﭽﻴﺪﻩ ﻣﻘﺎدیﺮ اوﻝﻴﻪ ﺑﺮاي ﭘﺎراﻣﺘﺮهﺎي ﻣﺠﻬﻮل ﻧﻴﺴﺖ‪.‬‬ ‫ﺵﺮط هﻢ ﺧﻄﻲ ﺑﺎ ﻓﻀﺎي ﺗﺼﻮیﺮ و ﻣﺪل ﺱﺮوآﺎر دارد‪ ،‬در ﺣﺎﻝﻴﻜﻪ ﺵﺮط هﻢ ﺻﻔﺤﻪ اي ﺗﻨﻬﺎ ﺑﺎ ﻓﻀﺎي ﺗﺼﻮیﺮ ﺱﺮوآﺎر دارد‪.‬‬ ‫اﺑﻌﺎد آﻮﭼﻚ ﻣﺎﺗﺮیﺲ ﻧﺮﻣﺎل‪ .‬ﺑﺎ اﻓﺰایﺶ ﻧﻘﺎط‪ ،‬و ﻣﺎﺗﺮیﺲ ﻧﺮﻣﺎل ﺑﺰرگ ﻧﻤﻲ ﺵﻮد‬

‫‪X = ( AT . A) −1 A.T L‬‬

‫[‬

‫‪45‬‬

‫)‪mathematical equations under photogrammetric conditions .... Safa Khazaei (winter, 2007‬‬

‫‪12‬‬