在大地测量中，求椭圆弧长时常常会碰到椭圆积分的问题，下面将其完整的表述出来
引入问题是求椭球的子午线的弧长
在经度为某一定值的子午线上，其微分弧长为：d S = M d B dS=MdB d S = M d B M = 1 κ ( B ) = a ( 1 − e 2 ) ( 1 − e 2 sin 2 B ) 3 / 2 M=\frac{1}{\kappa (B)}=\frac{a(1-e^2)}{(1-e^2\sin^2 B)^{3/2}} M = κ ( B ) 1 = ( 1 − e 2 sin 2 B ) 3 / 2 a ( 1 − e 2 ) B 1 B_1 B 1 B 2 B_2 B 2 s = ∫ B 1 B 2 M d B = a ( 1 − e 2 ) ∫ B 1 B 2 ( 1 − e 2 sin 2 B ) − 3 / 2 d B s=\int_{B_1}^{B_2}MdB=a(1-e^2)\int_{B_1}^{B_2}(1-e^2\sin^2 B)^{-3/2}dB s = ∫ B 1 B 2 M d B = a ( 1 − e 2 ) ∫ B 1 B 2 ( 1 − e 2 sin 2 B ) − 3 / 2 d B
设椭圆的参数方程为：{ x = a cos θ y = b sin θ \left\{
\begin{aligned}
&x=a\cos \theta\\
&y=b\sin \theta
\end{aligned}
\right. { x = a cos θ y = b sin θ θ \theta θ d s ds d s d s = x ′ 2 + y ′ 2 d θ = a 2 sin 2 θ + b 2 cos 2 θ d θ = a 1 − e 2 cos 2 θ d θ \begin{aligned}
ds=&\sqrt{x'^2+y'^2}d\theta \\
=&\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}d\theta\\
=&a\sqrt{1-e^2\cos^2\theta}d\theta
\end{aligned} d s = = = x ′ 2 + y ′ 2
d θ a 2 sin 2 θ + b 2 cos 2 θ
d θ a 1 − e 2 cos 2 θ
d θ ϕ = π / 2 − θ \phi=\pi/2-\theta ϕ = π / 2 − θ θ ∈ [ 0 , x ] \theta\in [0,x] θ ∈ [ 0 , x ] s = ∫ 0 x a 1 − e 2 cos 2 θ d θ = ∫ π 2 − x π 2 1 − e 2 sin 2 ϕ d ϕ \begin{aligned}
s=&\int_0^x a\sqrt{1-e^2\cos^2 \theta}d\theta \\
=&\int_{\frac{\pi}{2}-x}^{\frac{\pi}{2}} \sqrt{1-e^2\sin^2 \phi}d\phi
\end{aligned} s = = ∫ 0 x a 1 − e 2 cos 2 θ
d θ ∫ 2 π − x 2 π 1 − e 2 sin 2 ϕ
d ϕ
For expressing one argument:
α \alpha α k = sin α k=\sin\alpha k = sin α m = k 2 = sin 2 α m=k^2=\sin^2\alpha m = k 2 = sin 2 α
The incomplete elliptic integral of the first kind F is defined asF ( φ , k ) = F ( φ ∣ k 2 ) = F ( sin φ ; k ) = ∫ 0 φ d θ 1 − k 2 sin 2 θ . F(\varphi,k) = F\left(\varphi \,|\, k^2\right) = F(\sin \varphi ; k) = \int_0^\varphi \frac {d\theta}{\sqrt{1 - k^2 \sin^2 \theta}}. F ( φ , k ) = F ( φ ∣ k 2 ) = F ( sin φ ; k ) = ∫ 0 φ 1 − k 2 sin 2 θ
d θ . F ( φ ∣ m ) = ∫ 0 φ d θ 1 − m sin 2 θ F(\varphi|m)=\int_0^\varphi \frac {d\theta}{\sqrt{1 - m \sin^2 \theta}} F ( φ ∣ m ) = ∫ 0 φ 1 − m sin 2 θ
d θ
the complete elliptic integral of the first kind:K ( m ) = ∫ 0 π 2 d θ 1 − m sin 2 θ K(m)=\int_0^\frac{\pi}{2}\frac{d\theta}{\sqrt{1-m\sin^2\theta}} K ( m ) = ∫ 0 2 π 1 − m sin 2 θ
d θ
The incomplete elliptic integral of the second kind E in trigonometric form isE ( φ , k ) = ∫ 0 φ 1 − k 2 sin 2 θ d θ E(\varphi,k)=\int_0^\varphi\sqrt{1-k^2\sin^2\theta}d\theta E ( φ , k ) = ∫ 0 φ 1 − k 2 sin 2 θ
d θ E ( φ ∣ m ) = ∫ 0 φ 1 − m sin 2 θ d θ E(\varphi|m)=\int_0^\varphi\sqrt{1-m\sin^2\theta}d\theta E ( φ ∣ m ) = ∫ 0 φ 1 − m sin 2 θ
d θ φ \varphi φ φ = π / 2 − u \varphi=\pi/2-u φ = π / 2 − u
The incomplete elliptic integral of the third kind Π \Pi Π Π ( n ; φ ∣ m ) = ∫ 0 φ 1 1 − n sin 2 θ d θ 1 − m sin 2 θ \Pi(n;\varphi|m)=\int_0^\varphi\frac{1}{1-n\sin^2\theta}\frac{d\theta}{\sqrt{1-m\sin^2 \theta}} Π ( n ; φ ∣ m ) = ∫ 0 φ 1 − n sin 2 θ 1 1 − m sin 2 θ
d θ φ \varphi φ Π \Pi Π m ( φ ) = a ( 1 − e 2 ) Π ( e 2 ; φ ∣ e 2 ) m(\varphi)=a(1-e^2)\Pi(e^2;\varphi|e^2) m ( φ ) = a ( 1 − e 2 ) Π ( e 2 ; φ ∣ e 2 )
第一类椭圆积分的反函数
The The Jacobi elliptic functions are defined in terms of the integralu = ∫ 0 φ d θ 1 − m sin 2 φ u=\int_0^\varphi\frac{d\theta}{\sqrt{1-m\sin^2\varphi}} u = ∫ 0 φ 1 − m sin 2 φ
d θ s n ( u ) = sin φ , c n ( u ) = cos φ , d n ( u ) = 1 − m sin 2 φ sn(u)=\sin\varphi,\quad cn(u)=\cos\varphi,\quad dn(u)=\sqrt{1-m\sin^2\varphi} s n ( u ) = sin φ , c n ( u ) = cos φ , d n ( u ) = 1 − m sin 2 φ
matlab:
E ( φ ∣ k 2 ) = ( 1 − k 2 ) Π ( k 2 ; φ ∣ k 2 ) + k 2 sin φ cos φ 1 − k 2 sin 2 φ E(\varphi|k^2)=(1-k^2)\Pi(k^2;\varphi|k^2)+\frac{k^2\sin\varphi\cos\varphi}{\sqrt{1-k^2\sin^2\varphi}} E ( φ ∣ k 2 ) = ( 1 − k 2 ) Π ( k 2 ; φ ∣ k 2 ) + 1 − k 2 sin 2 φ
k 2 sin φ cos φ p r o o f . proof. p r o o f . ⟸ E ( φ ∣ k 2 ) − ( 1 − k 2 ) Π ( k 2 ; φ ∣ k 2 ) = k 2 sin φ cos φ 1 − k 2 sin 2 φ \impliedby E(\varphi|k^2)-(1-k^2)\Pi(k^2;\varphi|k^2)=\frac{k^2\sin\varphi\cos\varphi}{\sqrt{1-k^2\sin^2\varphi}} ⟸ E ( φ ∣ k 2 ) − ( 1 − k 2 ) Π ( k 2 ; φ ∣ k 2 ) = 1 − k 2 sin 2 φ
k 2 sin φ cos φ φ \varphi φ R i g h t = k 2 ( cos 2 φ − sin 2 φ ) ( 1 − k 2 sin 2 φ ) + k 4 sin 2 φ cos 2 φ ( 1 − k 2 sin 2 φ ) 3 / 2 = k 4 sin 4 φ − 2 k 2 sin 2 φ + k 2 ( 1 − k 2 sin 2 φ ) 3 / 2 = L e f t \begin{aligned}
Right=&\frac{k^2(\cos^2\varphi-\sin^2\varphi)(1-k^2\sin^2\varphi)+k^4\sin^2\varphi\cos^2\varphi}{(1-k^2\sin^2\varphi)^{3/2}}\\
=&\frac{k^4\sin^4\varphi-2k^2\sin^2\varphi+k^2}{(1-k^2\sin^2\varphi)^{3/2}}=Left
\end{aligned} R i g h t = = ( 1 − k 2 sin 2 φ ) 3 / 2 k 2 ( cos 2 φ − sin 2 φ ) ( 1 − k 2 sin 2 φ ) + k 4 sin 2 φ cos 2 φ ( 1 − k 2 sin 2 φ ) 3 / 2 k 4 sin 4 φ − 2 k 2 sin 2 φ + k 2 = L e f t φ = 0 \varphi=0 φ = 0
MatlabB , u B,u B , u
大地纬度表示s = a ( 1 − e 2 ) Π ( e 2 ; B ∣ e 2 ) = a ( 1 − e 2 ) ( e l l i p t i c P i ( e 2 , B , e 2 ) s=a(1-e^2)\Pi(e^2;B|e^2)=a(1-e^2)(ellipticPi(e^2,B,e^2) s = a ( 1 − e 2 ) Π ( e 2 ; B ∣ e 2 ) = a ( 1 − e 2 ) ( e l l i p t i c P i ( e 2 , B , e 2 )
归化纬度表示s = a ( E ( π 2 , e ) − E ( π 2 − u , e ) ) = a ( e l l i p t i c E ( e 2 ) − e l l i p t i c E ( p i / 2 − u , e 2 ) ) s=a(E(\frac{\pi}{2},e)-E(\frac{\pi}{2}-u,e))=a(ellipticE(e^2)-ellipticE(pi/2-u,e^2)) s = a ( E ( 2 π , e ) − E ( 2 π − u , e ) ) = a ( e l l i p t i c E ( e 2 ) − e l l i p t i c E ( p i / 2 − u , e 2 ) )
( 1 + x ) m = 1 + m x + m ( m − 1 ) 2 ! x 2 + ⋯ + m ( m − 1 ) ⋯ ( m − n + 1 ) n ! x n ( − 1 < x < 1 ) (1+x)^m=1+mx+\frac{m(m-1)}{2!}x^2+\cdots+\frac{m(m-1)\cdots(m-n+1)}{n!}x^n\quad(-1\lt x\lt 1) ( 1 + x ) m = 1 + m x + 2 ! m ( m − 1 ) x 2 + ⋯ + n ! m ( m − 1 ) ⋯ ( m − n + 1 ) x n ( − 1 < x < 1 ) 1 + x = 1 + 1 2 x − 1 2 ⋅ 4 x 2 + 1 ⋅ 3 2 ⋅ 4 ⋅ 6 x 3 − ⋯ + ( − 1 ) n ( 2 n − 3 ) ! ! ( 2 n ) ! ! x n + ⋯ ( − 1 ≤ x ≤ 1 ) \sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{2\cdot 4}x^2+\frac{1\cdot 3}{2\cdot 4\cdot 6}x^3-\cdots+(-1)^n\frac{(2n-3)!!}{(2n)!!}x^n+\cdots\quad (-1\le x\le1) 1 + x
= 1 + 2 1 x − 2 ⋅ 4 1 x 2 + 2 ⋅ 4 ⋅ 6 1 ⋅ 3 x 3 − ⋯ + ( − 1 ) n ( 2 n ) ! ! ( 2 n − 3 ) ! ! x n + ⋯ ( − 1 ≤ x ≤ 1 ) 1 1 + x = 1 − 1 2 x + 1 ⋅ 3 2 ⋅ 4 x 2 − 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 x 3 + ⋯ + ( − 1 ) n ( 2 n − 1 ) ! ! ( 2 n ) ! ! x n + ⋯ ( − 1 < x ≤ 1 ) \frac{1}{\sqrt{1+x}}=1-\frac{1}{2}x+\frac{1\cdot3}{2\cdot 4}x^2-\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^3+\cdots+(-1)^n\frac{(2n-1)!!}{(2n)!!}x^n+\cdots\quad(-1\lt x\le 1) 1 + x
1 = 1 − 2 1 x + 2 ⋅ 4 1 ⋅ 3 x 2 − 2 ⋅ 4 ⋅ 6 1 ⋅ 3 ⋅ 5 x 3 + ⋯ + ( − 1 ) n ( 2 n ) ! ! ( 2 n − 1 ) ! ! x n + ⋯ ( − 1 < x ≤ 1 )
Elliptic integral .wikipedia
matlab help documentation.
过家春，等；基于第二类椭圆积分的子午线弧长公式变换及解算(错误较多，有修订)
胡绍宗；椭圆积分的计算及应用