円筒の誤差伝搬
円筒の断面積は,外径をD,厚みをtとすると,
\( \Large \begin{eqnarray}S &=& \pi \left( \frac{D}{2} \right)^2- \pi \left( \frac{D}{2} -t \right)^2 \\
&=&
\pi \left( \frac{D}{2} \right)^2- \pi \left[ \left(\frac{D}{2} \right)^2-Dt + t^2 \right] \\
&=& \pi ( D-t )t \\
\end{eqnarray} \)
長さをL,質量をM,とすると体積Vと密度ρは,
\( \Large V = \pi ( D-t )t L \)
\( \Large \rho = \frac{M}{\pi ( D-t )t L} \)
となります.
誤差伝搬は,\( \Large y = f (x_a, x_b, x_c \cdots) \) とすると,
\( \Large \sigma_y \simeq \sqrt{ \left( \frac{ \partial f}{\partial x_a} \sigma_a \right)^2 + \left( \frac{ \partial f}{\partial x_b} \sigma_b \right)^2 + \left( \frac{ \partial f}{\partial x_c} \sigma_c \right)^2 + \cdots } \)
となるので,これをもとに計算していきます..
・円筒の体積
\( \Large \frac{ \partial V}{\partial D} = \pi tL = \pi ( D-t )t L \cdot \frac{1}{D-t} = V \cdot \frac{1}{D-t}\)
\( \Large \frac{ \partial V}{\partial L} = \pi ( D-t )t = \pi ( D-t )t L \cdot \frac{1}{L} = V \cdot \frac{1}{L} \)
\( \Large \frac{ \partial V}{\partial t} = \pi ( D-2t )L = \pi ( D-t )t L \cdot \frac{D-2t}{(D-t)t} = V \cdot \frac{D-2t}{(D-t)t} \)
となるので,
\( \Large \begin{eqnarray} \delta V &=& \sqrt{ \left( V \cdot \frac{1}{D-1} \delta_D \right)^2 + \left( V \cdot \frac{1}{L} \delta_L \right)^2+ \left( V \cdot \frac{D-2t}{(D-t)t} \delta_t \right)^2 } \\
&=& V \sqrt{ \left( \frac{1}{D-1} \delta_D \right)^2 + \left( \frac{1}{L} \delta_L \right)^2+ \left( \frac{D-2t}{(D-t)t} \delta_t \right)^2 } \\
&=& V \sqrt{ \left( \frac{D}{D-1} \frac{\delta_D}{D} \right)^2 + \left( \frac{\delta_L}{L} \right)^2+ \left( \frac{D-2t}{D-t} \frac{\delta_t}{t} \right)^2 } \\
\end{eqnarray} \)
となります.
D >> tのときは,
\( \Large \delta V \simeq V \sqrt{ \left( \frac{\delta_D}{D} \right)^2 + \left( \frac{\delta_L}{L} \right)^2+ \left(\frac{\delta_t}{t} \right)^2 } \)
となります.
・円筒の密度
\( \Large \rho = \frac{M}{\pi ( D-t )t L} \)
ですので,
\( \Large \frac{ \partial \rho}{\partial M} = \frac{1}{\pi ( D-t )t L} = \rho \cdot \frac{1}{M}\)
\( \Large \frac{ \partial \rho}{\partial L} = \frac{-M}{\pi ( D-t )t L^2} = \rho \cdot \frac{-1}{L}\)
\( \Large \begin{eqnarray} \frac{ \partial \rho}{\partial D} &=& \frac{M}{\pi L t} \frac{ \partial }{\partial D} \frac{1}{D-t} \\
&=& \frac{M}{\pi L t} \frac{-1}{(D-t)^2} \\
&=& \frac{M}{\pi ( D-t )t L} \frac{-1}{D-t} \\
&=& \rho \frac{-1}{D-t} \\
\end{eqnarray} \)
\( \Large \begin{eqnarray} \frac{ \partial \rho}{\partial t} &=& \frac{M}{\pi L} \frac{ \partial }{\partial t} \frac{1}{(D-t)t} \\
&=& \frac{M}{\pi L} \frac{ \partial }{\partial t} (Dt-t^2)^{-1} \\
&=& \frac{M}{\pi L} \frac{-(D-2t)}{ (Dt-t^2)^{2}} \\
&=& \frac{M}{\pi ( D-t )t L} \frac{-(D-2t)}{ (D-t)t} \\
&=& \rho \frac{-(D-2t)}{ (D-t)t} \\
\end{eqnarray} \)
となるので,
\( \Large \begin{eqnarray} \delta \rho &=& \sqrt{ \left( \rho \cdot \frac{1}{M} \delta_M \right)^2 + \left( \rho \cdot \frac{-1}{L} \delta_L \right)^2 + \left( \rho \cdot \frac{-1}{D-t} \delta_D \right)^2 + \left( \rho \cdot \frac{-(D-2t)}{(D-t)t} \delta_t \right)^2 } \\
&=& \rho \sqrt{ \left( \frac{\delta_M}{M} \right)^2 + \left( \frac{\delta_L}{L} \right)^2 + \left( \frac{D}{D-t} \frac{\delta_D}{D} \right)^2 + \left( \frac{D-2t}{D-t} \frac{\delta_t}{t} \right)^2} \\
\end{eqnarray} \)
となります.
D >> tのときは,
\( \Large \delta \rho = \rho \sqrt{ \left( \frac{\delta_M}{M} \right)^2 + \left( \frac{\delta_L}{L} \right)^2 + \left( \frac{\delta_D}{D} \right)^2 + \left( \frac{\delta_t}{t} \right)^2} \)
となります.