長岡係數を導出せよ

ゑ？

![fig-01.png](https://qiita-image-store.s3.ap-northeast-1.amazonaws.com/0/659995/026171d4-45f0-250a-3726-fb9816601d52.png)




$$ L = K\frac{{\mu _0 S N^2 }}{{h}} $$

$$ S=\pi a^2 $$

$$  dA\left( {z,\zeta ,\theta } \right) = \frac{{\mu _0 }}{{4\pi }}   \, \frac{{Jc \ d\zeta }}{{\tau \left( {\zeta  - z,\theta } \right)}} $$

$$ Jc = \frac{{I \, N}}{h} $$

$$ \tau \left( {x,\theta } \right) = \sqrt {x^2  + 4a^2 \sin ^2 \frac{\theta }{2}}  $$

```math
\begin{align*}
 \Phi c\left( {z,\zeta } \right) = 2\pi a\int_\alpha ^{2\pi  + \alpha } {dA\left( {z,\zeta ,\theta } \right)\cos \theta ds}  \\ 
  = 4\pi a\int_\psi ^{\pi  + \psi } {dA\left( {z,\zeta ,\theta } \right)\cos \theta ds}  \\ 
\end{align*}
```

$$ ds = a \, d\theta  $$

$$ \Phi \left( z \right) = \int_{\zeta  = 0}^{\zeta  = h } {\Phi c\left( {z,\zeta } \right) }  $$

$$ \Phi m = \int_0^h {\Phi \left( z \right)N\frac{{dz}}{h}} $$

$$ L = \frac{{\Phi m}}{I} $$

$  $

```math
\begin{align*}
 K &= \frac{h}{{\mu _0 SN^2 }}L = \frac{h}{{\mu _0 SN^2 }} \cdot \frac{{\Phi m}}{I} \\ 
  &= \frac{{h}}{{\mu _0 \pi a^2 N^2 I}}\int_0^h {\int_0^h {4\pi a\int_\psi ^{\pi  + \psi } {\frac{{\mu _0 }}{{4\pi }}\frac{{Jc\cos \theta }}{{\tau \left( {\zeta  - z,\theta } \right)}}\;ds} } d\zeta \frac{N}{h}dz}  \\ 
  &= \frac{{h}}{{\mu _0 \pi a^2 N^2 I}} \cdot \frac{{\mu _0 }}{{4\pi }}\int_0^h {\int_0^h {4\pi a\int_\psi ^{\pi  + \psi } {\frac{{I\;N}}{h}\frac{{a\;\cos \theta }}{{\tau \left( {\zeta  - z,\theta } \right)}}\;d\theta } \;} d\zeta \frac{N}{h}dz}  \\ 
  &= \frac{1}{{\pi h}}\int_0^h {\int_0^h {\int_\psi ^{\pi  + \psi } {\frac{{\cos \theta }}{{\tau \left( {\zeta  - z,\theta } \right)}}\;d\theta } \;} d\zeta dz}  \\ 
\end{align*}
```
$  $

```math
\begin{align*}

\eqalign{
  & \phi  + \frac{\pi }
{2} = \frac{\theta }
{2}  \cr 
  & \theta  = 2\phi  + \pi   \cr 
  & d\theta  = 2d\phi   \cr 
  & \sin \frac{\theta }
{2} = \cos \phi   \cr 
  & \tau \left( {x,\theta } \right) = \sqrt {x^2  + 4a^2 \sin ^2 \frac{\theta }
{2}}  = \sqrt {x^2  + 4a^2 \cos ^2 \phi }  \cr} 

\end{align*} 
```



```math
\begin{align*}

\eqalign{
  K &= \frac{1}
{{\pi h}}\int_0^h {\int_0^h {\int_{n\pi }^{\left( {n + 1} \right)\pi } {\frac{{\cos \theta }}
{{\tau \left( {\zeta  - z,\theta } \right)}}\;d\theta } \;} d\zeta dz}   \cr 
  &  = \frac{1}
{{\pi h}}\int_{n\pi }^{\left( {n + 1} \right)\pi } {\cos \theta \int_0^h {\int_0^h {\frac{1}
{{\sqrt {\left( {\zeta  - z} \right)^2  + 4a^2 \sin ^2 \frac{\theta }
{2}} }}} d\zeta dz} d\theta }   \cr 
  &  = \frac{1}
{{\pi h}}\int_{n\pi  = 2\phi  + \pi }^{\left( {n + 1} \right)\pi  = 2\phi  + \pi } {\left( { - \cos 2\phi } \right)\int_0^h {\int_0^h {\frac{1}
{{\sqrt {\left( {\zeta  - z} \right)^2  + 4a^2 \cos ^2 \phi } }}} d\zeta dz} \;2d\phi }   \cr 
  &  =  - \frac{2}
{{\pi h}}\int_{\phi  = \left( {n - 1} \right)\frac{\pi }
{2}}^{\phi  = n\frac{\pi }
{2}} {\cos 2\phi \int_0^h {\int_0^h {\frac{1}
{{\sqrt {\left( {\zeta  - z} \right)^2  + 4a^2 \cos ^2 \phi } }}} d\zeta dz} d\phi }   \cr 
  &  =  - \frac{2}
{{\pi h}}\int_0^{\frac{\pi }
{2}} {\cos 2\phi \int_0^h {\int_{ - z}^{h - z} {\frac{1}
{{\sqrt {v^2  + 4a^2 \cos ^2 \phi } }}dv} dz} d\phi }  \cr} 

\end{align*} 
```

```math
\begin{align*}

\eqalign{
  & fa = \frac{1}
{{\sqrt {v^2  + 4a^2 \cos ^2 \phi } }}  \cr 
  & \int_0^h {\int_{ - z}^{h - z} {fa\;dv} dz}  = \int_{}^h {\int_{}^{h - z} {fa\;dv} dz}  - \int_{}^h {\int_{}^{ - z} {fa\;dv} dz}  - \int_{}^0 {\int_{}^{h - z} {fa\;dv} dz}  + \int_{}^0 {\int_{}^{ - z} {fa\;dv} dz}   \cr 
  &  =  - \int_{}^0 {\int {fa\;dv^2 } }  + \int_{}^{ - h} {\int {fa\;dv^2 } }  + \int_{}^h {\int {fa\;dv^2 } }  - \int_{}^0 {\int {fa\;dv^2 } }   \cr 
  &  = \int_{}^h {\int {fa\;dv^2 } }  + \int_{}^{ - h} {\int {fa\;dv^2 } }  - 2\int_{}^0 {\int {fa\;dv^2 } }  \cr} 

\end{align*} 
```

```math
\begin{align*}

\eqalign{
  K & =  - \frac{2}
{{\pi h}}\int_0^{\frac{\pi }
{2}} {\cos 2\phi \int_0^h {\int_{ - z}^{h - z} {fa\;dv} dz} d\phi }   \cr 
  &  =  - \frac{2}
{{\pi h}}\int_0^{\frac{\pi }
{2}} {\cos 2\phi \left( {\int_{}^h {\int {fa\;dv^2 } }  + \int_{}^{ - h} {\int {fa\;dv^2 } }  - 2\int_{}^0 {\int {fa\;dv^2 } } } \right)d\phi }   \cr 
  &  =  - \frac{2}
{{\pi h}}\int_0^{\frac{\pi }
{2}} {\cos 2\phi \left( {\int_{}^h {\int {\frac{1}
{{\sqrt {v^2  + 4a^2 \cos ^2 \phi } }}dv^2 } }  + \int_{}^{ - h} {\int {\frac{1}
{{\sqrt {v^2  + 4a^2 \cos ^2 \phi } }}dv^2 } }  - 2\int_{}^0 {\int {\frac{1}
{{\sqrt {v^2  + 4a^2 \cos ^2 \phi } }}dv^2 } } } \right)d\phi }  \cr} 
\end{align*} 
```

```math
\begin{align*}

\eqalign{
  & Fa\left( {t,b} \right) = \int_{}^t {\int {\frac{1}
{{\sqrt {x^2  + b^2 } }}dx^2 } }  = \int_{}^t {\ln \left( {x + \sqrt {x^2  + b^2 } } \right)dx}  = t\ln \left( {t + \sqrt {t^2  + b^2 } } \right) - \sqrt {t^2  + b^2 }   \cr 
  & Fb\left( t \right) =  - \int_0^{\frac{\pi }
{2}} {\cos 2\phi \;Fa\left( {t,2a\cos \phi } \right)d\phi }  \cr} 

\end{align*} 
```

```math
\begin{align*}

\eqalign{
  K & =  - \frac{2}
{{\pi h}}\int_0^{\frac{\pi }
{2}} {\cos 2\phi \left\{ {Fa\left( {h,2a\cos \phi } \right) + Fa\left( { - h,2a\cos \phi } \right) - 2Fa\left( {0,2a\cos \phi } \right)} \right\}d\phi }   \cr 
  &  = \frac{2}
{{\pi h}}\left\{ {Fb\left( h \right) + Fb\left( { - h} \right) + 2Fb\left( 0 \right)} \right\} \cr} 

\end{align*} 
```

```math
\begin{align*}

\eqalign{
  Fb\left( x \right) & =  - \int_0^{\frac{\pi }
{2}} {\cos 2\phi \;Fa\left( {x,2a\cos \phi } \right)d\phi }   \cr 
  &  =  - \int_0^{\frac{\pi }
{2}} {\cos 2\phi \;\left\{ {x\ln \left( {x + \sqrt {x^2  + 4a^2 \cos ^2 \phi } } \right) - \sqrt {x^2  + 4a^2 \cos ^2 \phi } } \right\}d\phi }   \cr 
  &  =  - \left[ {\sin \phi \cos \phi Fa\left( {x,2a\cos \phi } \right)} \right]_0^{\frac{\pi }
{2}}  + \int_0^{\frac{\pi }
{2}} {\left( {\sin \phi \cos \phi } \right)\tan \phi \left( {\sqrt {x^2  + 4a^2 \cos ^2 \phi }  - x} \right)d\phi }   \cr 
  &  = \int_0^{\frac{\pi }
{2}} {\sin ^2 \phi \left( {\sqrt {x^2  + 4a^2 \cos ^2 \phi }  - x} \right)d\phi }  \cr} 

\end{align*} 
```
$$
Fb\left( x \right) + Fb\left( { - x} \right) = 2\int_0^{\frac{\pi }
{2}} {\sin ^2 \phi \sqrt {x^2  + 4a^2 \cos ^2 \phi } d\phi } 
$$

$$
Fb\left( 0 \right) =  - \int_0^{\frac{\pi }
{2}} {\sin ^2 \phi \left( {2a\cos \phi } \right)d\phi }  =  - \frac{{2a}}
{3}
$$


```math
\begin{align*}

\eqalign{
  & \sqrt {h^2  + 4a^2 \cos ^2 \phi }  = \sqrt {h^2  + 4a^2  - 4a^2 \sin ^2 \phi }   \cr 
  &  = \sqrt {h^2  + 4a^2 } \sqrt {1 - \frac{{4a^2 }}
{{h^2  + 4a^2 }}\sin ^2 \phi }  = 2ak^{ - 1} \sqrt {1 - k^2 \sin ^2 \phi }   \cr 
  & k = \frac{{2a}}
{{\sqrt {h^2  + 4a^2 } }} = \sqrt {1 + \left( {\frac{h}
{{2a}}} \right)^2 } ^{ - 1} = \cos{\left( \arctan{\frac{2a}{{h}}} \right)}    \cr 
  & 1 + \left( {\frac{h}
{{2a}}} \right)^2  = k^{ - 2}   \cr 
  & \frac{h}
{{2a}} = \sqrt {k^{ - 2}  - 1}  = \frac{{\sqrt {1 - k^2 } }}
{k} \cr} 

\end{align*} 
```

```math
\begin{align*}

\eqalign{
  K & = \frac{2}
{{\pi h}}\left\{ {Fb\left( h \right) + Fb\left( { - h} \right) + 2Fb\left( 0 \right)} \right\}  \cr 
  &  = \frac{2}
{{\pi h}}\left( {2\int_0^{\frac{\pi }
{2}} {\sin ^2 \phi \sqrt {x^2  + 4a^2 \cos ^2 \phi } d\phi }  - \frac{{4a}}
{3}} \right)  \cr 
  &  = \frac{2}
{{\pi h}}\left( {4ak^{ - 1} \int_0^{\frac{\pi }
{2}} {\sin ^2 \phi \sqrt {1 - k^2 \sin ^2 \phi } d\phi }  - \frac{{4a}}
{3}} \right)  \cr 
  &  = \frac{{8a}}
{{3\pi h}}\left( {3k^{ - 1} \int_0^{\frac{\pi }
{2}} {\sin ^2 \phi \sqrt {1 - k^2 \sin ^2 \phi } d\phi }  - 1} \right)  \cr 
  &  = \frac{{4k}}
{{3\pi \sqrt {1 - k^2 } }}\left( {3k^{ - 1} \int_0^{\frac{\pi }
{2}} {\sin ^2 \phi \sqrt {1 - k^2 \sin ^2 \phi } d\phi }  - 1} \right)  \cr 
  &  = \frac{4}
{{3\pi \sqrt {1 - k^2 } }}\left( {3\int_0^{\frac{\pi }
{2}} {\sin ^2 \phi \sqrt {1 - k^2 \sin ^2 \phi } d\phi }  - k} \right) \cr} 

\end{align*} 
```

```math
\begin{align*}

\eqalign{
  & p = \sqrt {1 - k^2 \sin ^2 \phi }   \cr 
  & k^2 \sin ^2 \phi  = 1 - p^2 ,\;\sin ^2 \phi  = \frac{{1 - p^2 }}
{{k^2 }} = k^{ - 2}  - k^{ - 2} p^2   \cr 
  & \frac{d}
{{d\phi }}p =  - k^2 p^{ - 1} \sin \phi \cos \phi  \cr} 

\end{align*} 
```

```math
\begin{align*}

\eqalign{
  \frac{d}
{{d\phi }}p\sin \phi \cos \phi  & = \left( { - k^2 p^{ - 1} \sin \phi \cos \phi } \right)\sin \phi \cos \phi  + p\left( {1 - 2\sin ^2 \phi } \right)  \cr 
  &  = \left( {1 - 2\sin ^2 \phi } \right)p - p^{ - 1} k^2 \sin ^2 \phi \cos ^2 \phi   \cr 
  &  = \left( {1 - 2\sin ^2 \phi } \right)p - \left( {1 - p^2 } \right)\left( {1 - \sin ^2 \phi } \right)p^{ - 1}   \cr 
  &  = \left( {1 - 2\sin ^2 \phi } \right)p - \left( {p^{ - 1}  - p} \right)\left( {1 - \sin ^2 \phi } \right)  \cr 
  &  = \left( {1 - 2\sin ^2 \phi } \right)p - \left( {1 - \sin ^2 \phi } \right)p^{ - 1}  + \left( {1 - \sin ^2 \phi } \right)p  \cr 
  &  = \left( {2 - 3\sin ^2 \phi } \right)p - \left( {1 - k^{ - 2}  + k^{ - 2} p^2 } \right)p^{ - 1}   \cr 
  &  = 2p - 3p\sin ^2 \phi  - \left( {1 - k^{ - 2} } \right)p^{ - 1}  - k^{ - 2} p \cr} 

\end{align*} 
```

```math
\begin{align*}

\eqalign{
  & 3p\sin ^2 \phi  = \left( {k^{ - 2}  - 1} \right)p^{ - 1}  - \left( {k^{ - 2}  - 2} \right)p - \frac{d}
{{d\phi }}p\sin \phi \cos \phi   \cr 
  & 3\int_0^{\frac{\pi }
{2}} {\sin ^2 \phi \sqrt {1 - k^2 \sin ^2 \phi } d\phi }  = \left( {k^{ - 2}  - 1} \right)K\left( k \right) - \left( {k^{ - 2}  - 2} \right)E\left( k \right) \cr} 

\end{align*} 
```



```math
\begin{align*}

\eqalign{
  K & = \frac{4}
{{3\pi \sqrt {1 - k^2 } }}\left( {3\int_0^{\frac{\pi }
{2}} {\sin ^2 \phi \sqrt {1 - k^2 \sin ^2 \phi } d\phi }  - k} \right)  \cr 
  &  = \frac{4}
{{3\pi \sqrt {1 - k^2 } }}\left\{ {\left( {k^{ - 2}  - 1} \right)K\left( k \right) - \left( {k^{ - 2}  - 2} \right)E\left( k \right) - k} \right\}  \cr 
  &  = \frac{4}
{{3\pi \sqrt {1 - k^2 } }}\left\{ {\left( {\frac{1}
{{k^2 }} - 1} \right)K\left( k \right) - \left( {\frac{1}
{{k^2 }} - 2} \right)E\left( k \right) - k} \right\} \cr} 

\end{align*} 
```


ιょ ゃ ぃ b