LLA(经纬高)坐标转换成ENU(东北天)坐标的详细推导

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这是一篇经纬高(LLA)坐标转东北天坐标(ENU)的详细推导,并给出近似转换的过程和结果

参考资料:


https://blog.csdn.net/qq_34213260/article/details/109133847

在这里插入图片描述

ECEF坐标系和ENU坐标系之间的关系如上图所示,各坐标系的定义在此不再赘述。

LLA坐标:Lat、Lon、Alt坐标,表示在ECEF坐标系下,其中Lat和Lon用角度(



^。






























)或弧度(rad)表示,Alt用米(m)表示;

ENU坐标:东北天坐标系下的坐标,该坐标系的原点需要指定,原点向东为x轴,原点向北为y轴,原点指向天为z轴,构成右手直角坐标系,该系下的坐标表示都是米制单位;


LLA坐标转换成ENU坐标的过程主要分为两步:LLA->ECEF->ENU

其中主要参数的定义如下:

在这里插入图片描述





b

=

6356752.3142

b=6356752.3142






b




=








6


3


5


6


7


5


2


.


3


1


4


2











e

2

=

f

(

2

f

)

e^2=f(2-f)







e










2











=








f


(


2













f


)











N

=

a

1

e

2

s

i

n

2

(

l

a

t

)

N=\frac{a}{\sqrt{1-e^2sin^2(lat)}}






N




=



























1










e










2









s


i



n










2









(


l


a


t


)




































a

























其中b为椭球短半径,e为椭球偏心率


(1)LLA->ECEF:(无需给定起点,一对一转换)


从最上面一幅图中可以轻易得到同一点在这两个坐标系之间的转换关系,如下:





X

e

c

e

f

=

(

N

+

a

l

t

)

c

o

s

(

l

a

t

)

c

o

s

(

l

o

n

)

X_{ecef}=(N+alt)cos(lat)cos(lon)







X











e


c


e


f





















=








(


N




+








a


l


t


)


c


o


s


(


l


a


t


)


c


o


s


(


l


o


n


)











Y

e

c

e

f

=

(

N

+

a

l

t

)

c

o

s

(

l

a

t

)

s

i

n

(

l

o

n

)

Y_{ecef}=(N+alt)cos(lat)sin(lon)







Y











e


c


e


f





















=








(


N




+








a


l


t


)


c


o


s


(


l


a


t


)


s


i


n


(


l


o


n


)











Z

e

c

e

f

=

(

N

(

1

e

2

)

+

a

l

t

)

s

i

n

(

l

a

t

)

Z_{ecef}=(N(1-e^2)+alt)sin(lat)







Z











e


c


e


f





















=








(


N


(


1














e










2









)




+








a


l


t


)


s


i


n


(


l


a


t


)








(2)ECEF->ENU:(需要给定起点)


假设ENU坐标系的

坐标原点

对应的ECEF坐标和LLA坐标分别为:





O

e

c

e

f

=

(

X

0

,

Y

0

,

Z

0

)

O_{ecef}=(X_0,Y_0,Z_0)







O











e


c


e


f





















=








(



X










0


















,





Y










0


















,





Z










0


















)











O

l

l

a

=

(

l

a

t

0

,

l

o

n

0

,

a

l

t

0

)

O_{lla}=(lat_0,lon_0,alt_0)







O











l


l


a





















=








(


l


a



t










0


















,




l


o



n










0


















,




a


l



t










0


















)







假设空间中有另一点P,且:





P

e

c

e

f

=

(

X

,

Y

,

Z

)

P_{ecef}=(X,Y,Z)







P











e


c


e


f





















=








(


X


,




Y


,




Z


)







则P在ENU系下的坐标为:





P

e

n

u

=

R

E

C

E

F

E

N

U

(

P

e

c

e

f

O

e

c

e

f

)

(1)

P_{enu}=R_{ECEF}^{ENU}(P_{ecef}-O_{ecef})\tag{1}







P











e


n


u





















=









R











E


C


E


F










E


N


U



















(



P











e


c


e


f































O











e


c


e


f



















)







(



1



)








其中只有



R

E

C

E

F

E

N

U

R_{ECEF}^{ENU}







R











E


C


E


F










E


N


U






















是未知的,下面对其进行推导:




R

E

C

E

F

E

N

U

R_{ECEF}^{ENU}







R











E


C


E


F










E


N


U






















可以分解为如下三个旋转过程:

ECEF–>绕



Z

e

c

e

f

Z_{ecef}







Z











e


c


e


f






















轴逆时针旋转



λ

\lambda






λ





(经度)–>绕



Y

e

c

e

f

Y_{ecef}^{‘}







Y











e


c


e


f





















































轴逆时针旋转(



90

ϕ

90-\phi






9


0













ϕ





)(



ϕ

:

\phi:






ϕ




:





纬度)–>绕



Z

e

c

e

f

Z_{ecef}^{”}







Z











e


c


e


f
























































逆时针旋转90–>ENU

上述问题是典型的空间中点不动,坐标系转动的问题,按顺序记三次旋转分别为



R

1

R

2

R

3

R_1、R_2、R_3







R










1






















R










2






















R










3





















,则:





R

1

=

[

c

o

s

(

λ

)

s

i

n

(

λ

)

0

s

i

n

(

λ

)

c

o

s

(

λ

)

0

0

0

1

]

R_1= \begin{bmatrix} cos(\lambda) & sin(\lambda) & 0 \\ -sin(\lambda) & cos(\lambda) & 0 \\ 0 & 0 & 1 \end{bmatrix}







R










1




















=
























































c


o


s


(


λ


)











s


i


n


(


λ


)








0





























s


i


n


(


λ


)








c


o


s


(


λ


)








0





























0








0








1


































































R

2

=

[

s

i

n

(

ϕ

)

0

c

o

s

(

ϕ

)

0

1

0

c

o

s

(

ϕ

)

0

s

i

n

(

ϕ

)

]

R_2= \begin{bmatrix} sin(\phi) & 0 & -cos(\phi) \\ 0 & 1 & 0 \\ cos(\phi) & 0 & sin(\phi) \end{bmatrix}







R










2




















=
























































s


i


n


(


ϕ


)








0








c


o


s


(


ϕ


)





























0








1








0
































c


o


s


(


ϕ


)








0








s


i


n


(


ϕ


)


































































R

3

=

[

0

1

0

1

0

0

0

0

1

]

R_3= \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}







R










3




















=
























































0











1








0





























1








0








0





























0








0








1






























































则:





R

E

C

E

F

E

N

U

=

R

3

R

2

R

1

=

[

s

i

n

(

λ

)

c

o

s

(

λ

)

0

s

i

n

(

ϕ

)

c

o

s

(

λ

)

s

i

n

(

ϕ

)

s

i

n

(

λ

)

c

o

s

(

ϕ

)

c

o

s

(

ϕ

)

c

o

s

(

λ

)

c

o

s

(

ϕ

)

s

i

n

(

λ

)

s

i

n

(

ϕ

)

]

R_{ECEF}^{ENU}=R_3*R_2*R_1= \begin{bmatrix} -sin(\lambda) & cos(\lambda) & 0 \\ -sin(\phi)cos(\lambda) & -sin(\phi)sin(\lambda) & cos(\phi) \\ cos(\phi)cos(\lambda) & cos(\phi)sin(\lambda) & sin(\phi) \end{bmatrix}







R











E


C


E


F










E


N


U





















=









R










3






























R










2






























R










1




















=



























































s


i


n


(


λ


)











s


i


n


(


ϕ


)


c


o


s


(


λ


)








c


o


s


(


ϕ


)


c


o


s


(


λ


)





























c


o


s


(


λ


)











s


i


n


(


ϕ


)


s


i


n


(


λ


)








c


o


s


(


ϕ


)


s


i


n


(


λ


)





























0








c


o


s


(


ϕ


)








s


i


n


(


ϕ


)






























































将上式带入到公式(1)中可以得到:





P

e

n

u

=

R

E

C

E

F

E

N

U

(

P

e

c

e

f

O

e

c

e

f

)

=

[

s

i

n

(

l

o

n

0

)

c

o

s

(

l

o

n

0

)

0

s

i

n

(

l

a

t

0

)

c

o

s

(

l

o

n

0

)

s

i

n

(

l

a

t

0

)

s

i

n

(

l

o

n

0

)

c

o

s

(

l

a

t

0

)

c

o

s

(

l

a

t

0

)

c

o

s

(

l

o

n

0

)

c

o

s

(

l

a

t

0

)

s

i

n

(

l

o

n

0

)

s

i

n

(

l

a

t

0

)

]

[

X

X

0

Y

Y

0

Z

Z

0

]

(2)

P_{enu}=R_{ECEF}^{ENU}(P_{ecef}-O_{ecef})= \begin{bmatrix} -sin(lon_0) & cos(lon_0) & 0 \\ -sin(lat_0)cos(lon_0) & -sin(lat_0)sin(lon_0) & cos(lat_0) \\ cos(lat_0)cos(lon_0) & cos(lat_0)sin(lon_0) & sin(lat_0) \end{bmatrix}* \begin{bmatrix} X-X_0 \\ Y-Y_0 \\ Z-Z_0 \end{bmatrix}\tag{2}







P











e


n


u





















=









R











E


C


E


F










E


N


U



















(



P











e


c


e


f































O











e


c


e


f



















)




=



























































s


i


n


(


l


o



n










0


















)











s


i


n


(


l


a



t










0


















)


c


o


s


(


l


o



n










0


















)








c


o


s


(


l


a



t










0


















)


c


o


s


(


l


o



n










0


















)





























c


o


s


(


l


o



n










0


















)











s


i


n


(


l


a



t










0


















)


s


i


n


(


l


o



n










0


















)








c


o


s


(


l


a



t










0


















)


s


i


n


(


l


o



n










0


















)





























0








c


o


s


(


l


a



t










0


















)








s


i


n


(


l


a



t










0


















)




















































































































X










X










0
























Y










Y










0
























Z










Z










0














































































(



2



)









至此完成了LLA坐标到ENU坐标的转换,且无近似,下面对公式(2)进行近似推导


由于





X

=

(

N

+

a

l

t

)

c

o

s

(

l

a

t

)

c

o

s

(

l

o

n

)

X=(N+alt)cos(lat)cos(lon)






X




=








(


N




+








a


l


t


)


c


o


s


(


l


a


t


)


c


o


s


(


l


o


n


)











Y

=

(

N

+

a

l

t

)

c

o

s

(

l

a

t

)

s

i

n

(

l

o

n

)

Y=(N+alt)cos(lat)sin(lon)






Y




=








(


N




+








a


l


t


)


c


o


s


(


l


a


t


)


s


i


n


(


l


o


n


)











Z

=

(

N

(

1

e

2

)

+

a

l

t

)

s

i

n

(

l

a

t

)

Z=(N(1-e^2)+alt)sin(lat)






Z




=








(


N


(


1














e










2









)




+








a


l


t


)


s


i


n


(


l


a


t


)











X

0

=

(

N

0

+

a

l

t

0

)

c

o

s

(

l

a

t

0

)

c

o

s

(

l

o

n

0

)

X_0=(N_0+alt_0)cos(lat_0)cos(lon_0)







X










0




















=








(



N










0




















+








a


l



t










0


















)


c


o


s


(


l


a



t










0


















)


c


o


s


(


l


o



n










0


















)











Y

0

=

(

N

0

+

a

l

t

0

)

c

o

s

(

l

a

t

0

)

s

i

n

(

l

o

n

0

)

Y_0=(N_0+alt_0)cos(lat_0)sin(lon_0)







Y










0




















=








(



N










0




















+








a


l



t










0


















)


c


o


s


(


l


a



t










0


















)


s


i


n


(


l


o



n










0


















)











Z

0

=

(

N

0

(

1

e

2

)

+

a

l

t

0

)

s

i

n

(

l

a

t

0

)

Z_0=(N_0(1-e^2)+alt_0)sin(lat_0)







Z










0




















=








(



N










0


















(


1














e










2









)




+








a


l



t










0


















)


s


i


n


(


l


a



t










0


















)







所以





X

X

0

=

(

N

+

a

l

t

)

c

o

s

(

l

a

t

)

c

o

s

(

l

o

n

)

(

N

0

+

a

l

t

0

)

c

o

s

(

l

a

t

0

)

c

o

s

(

l

o

n

0

)

=

(

N

+

a

l

t

)

c

o

s

(

l

a

t

)

c

o

s

(

l

o

n

)

(

N

+

a

l

t

0

)

c

o

s

(

l

a

t

0

)

c

o

s

(

l

o

n

0

)

=

(

N

+

a

l

t

0

+

δ

a

l

t

)

c

o

s

(

l

a

t

0

+

δ

l

a

t

)

c

o

s

(

l

o

n

0

+

δ

l

o

n

)

(

N

+

a

l

t

0

)

c

o

s

(

l

a

t

0

)

c

o

s

(

l

o

n

0

)

=

(

N

+

a

l

t

0

)

c

o

s

(

l

a

t

0

+

δ

l

a

t

)

c

o

s

(

l

o

n

0

+

δ

l

o

n

)

(

N

+

a

l

t

0

)

c

o

s

(

l

a

t

0

)

c

o

s

(

l

o

n

0

)

+

δ

a

l

t

c

o

s

(

l

a

t

0

+

δ

l

a

t

)

c

o

s

(

l

o

n

0

+

δ

l

o

n

)

=

(

N

+

a

l

t

0

)

[

c

o

s

(

l

a

t

0

+

δ

l

a

t

)

c

o

s

(

l

o

n

0

+

δ

l

o

n

)

c

o

s

(

l

a

t

0

)

c

o

s

(

l

o

n

0

)

]

+

δ

a

l

t

c

o

s

(

l

a

t

0

+

δ

l

a

t

)

c

o

s

(

l

o

n

0

+

δ

l

o

n

)

=

(

N

+

a

l

t

0

)

[

(

c

o

s

(

l

a

t

0

)

c

o

s

(

δ

l

a

t

)

s

i

n

(

l

a

t

0

)

s

i

n

(

δ

l

a

t

)

)

(

c

o

s

(

l

o

n

0

)

c

o

s

(

δ

l

o

n

)

s

i

n

(

l

o

n

0

)

s

i

n

(

δ

l

o

n

)

)

c

o

s

(

l

a

t

0

)

c

o

s

(

l

o

n

0

)

]

+

δ

a

l

t

c

o

s

(

l

a

t

0

+

δ

l

a

t

)

c

o

s

(

l

o

n

0

+

δ

l

o

n

)

=

(

N

+

a

l

t

0

)

(

c

o

s

(

l

a

t

0

)

s

i

n

(

l

o

n

0

)

c

o

s

(

δ

l

a

t

)

s

i

n

(

δ

l

o

n

)

s

i

n

(

l

a

t

0

)

c

o

s

(

l

o

n

0

)

s

i

n

(

δ

l

a

t

)

c

o

s

(

δ

l

o

n

)

)

+

δ

a

l

t

c

o

s

(

l

a

t

0

+

δ

l

a

t

)

c

o

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\begin{aligned} X-X_0 &= (N+alt)cos(lat)cos(lon)-(N_0+alt_0)cos(lat_0)cos(lon_0)\\ &= (N+alt)cos(lat)cos(lon)-(N+alt_0)cos(lat_0)cos(lon_0)\\ &= (N+alt_0+\delta{alt})cos(lat_0+\delta{lat})cos(lon_0+\delta{lon})-(N+alt_0)cos(lat_0)cos(lon_0)\\ &= (N+alt_0)cos(lat_0+\delta{lat})cos(lon_0+\delta{lon})-(N+alt_0)cos(lat_0)cos(lon_0)+\delta{alt}cos(lat_0+\delta{lat})cos(lon_0+\delta{lon})\\ &= (N+alt_0)\left[cos(lat_0+\delta{lat})cos(lon_0+\delta{lon})-cos(lat_0)cos(lon_0) \right]+\delta{alt}cos(lat_0+\delta{lat})cos(lon_0+\delta{lon})\\ &= (N+alt_0)\left[(cos(lat_0)cos(\delta{lat})-sin(lat_0)sin(\delta{lat}))(cos(lon_0)cos(\delta{lon})-sin(lon_0)sin(\delta{lon}))-cos(lat_0)cos(lon_0) \right]+\delta{alt}cos(lat_0+\delta{lat})cos(lon_0+\delta{lon})\\ &= (N+alt_0)(-cos(lat_0)sin(lon_0)cos(\delta{lat})sin(\delta{lon})-sin(lat_0)cos(lon_0)sin(\delta{lat})cos(\delta{lon}))+\delta{alt}cos(lat_0+\delta{lat})cos(lon_0+\delta{lon})\\ &= (N+alt_0)(-cos(lat_0)sin(lon_0)*\delta{lon}-sin(lat_0)cos(lon_0)*\delta{lat})+\delta{alt}cos(lat_0+\delta{lat})cos(lon_0+\delta{lon})\\ &= (N+alt_0)(-cos(lat_0)sin(lon_0)*\delta{lon}-sin(lat_0)cos(lon_0)*\delta{lat})+\delta{alt}\left[(cos(lat_0)-sin(lat_0)*\delta{lat})(cos(lon_0)-sin(lon_0)*\delta{lon})\right]\\ &= (N+alt_0)(-cos(lat_0)sin(lon_0)*\delta{lon}-sin(lat_0)cos(lon_0)*\delta{lat})+\delta{alt}\left[cos(lat_0)cos(lon_0)-cos(lat_0)sin(lon_0)*\delta{lon}-cos(lon_0)sin(lat_0)*\delta{lat} \right]\\ &= (N+alt_0)(-cos(lat_0)sin(lon_0)*\delta{lon}-sin(lat_0)cos(lon_0)*\delta{lat})+\delta{alt}cos(lat_0)cos(lon_0) \end{aligned}
















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Y-Y_0=(N+alt_0)(\delta{lon}cos(lat_0)cos(lon_0)-\delta{lat}sin(lat_0)sin(lon_0))+\delta{alt}cos(lat_0)sin(lon_0)






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\begin{aligned} e &= -sin(lon_0)*(X-X_0)+cos(lon_0)*(Y-Y_0)\\ &= -sin(lon_0)((N+alt_0)(-cos(lat_0)sin(lon_0)*\delta{lon}-sin(lat_0)cos(lon_0)*\delta{lat})+\delta{alt}cos(lat_0)cos(lon_0))+cos(lon_0)((N+alt_0)(\delta{lon}cos(lat_0)cos(lon_0)-\delta{lat}sin(lat_0)sin(lon_0))+\delta{alt}cos(lat_0)sin(lon_0))\\ &= (N+alt_0)(sin(lon_0)cos(lat_0)sin(lon_0)*\delta{lon})+(N+alt_0)(cos(lat_0)cos^2(lon_0)*\delta{lon})\\ &= (N+alt_0)*\delta{lon}cos(lat_0)(sin^2(lon_0)+cos^2(lon_0))\\ &= (N+alt_0)*\delta{lon}cos(lat_0)\\ &\approx a*\delta{lon}cos(lat_0) \end{aligned}
















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类似的,可以得到n、u坐标,最终得到如下的近似结果:





P

e

n

u

=

[

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]

P_{enu}= \begin{bmatrix} e\\n\\u \end{bmatrix}= \begin{bmatrix} a*\delta{lon}cos(lat_0)\\a*\delta{lat}\\\delta{alt} \end{bmatrix}







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