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\begin{eqnarray}PGA=|a_X(t)|_{max}\end{eqnarray}

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\begin{eqnarray}PGA=|\sqrt[]{a_X^2(t)+a_Y^2(t)}|_{max}\end{eqnarray}

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\begin{eqnarray}PGA=|\sqrt[]{a_X^2(t)+a_Y^2(t)+a_Z^2(t)}|_{max}\end{eqnarray}

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\begin{eqnarray}I_{JMA}=2log(a_0)+0.94\end{eqnarray}

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\begin{eqnarray}\int_0^{T_d}w(t,a)dt \geq 0.3 \end{eqnarray}

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\begin{eqnarray}w_T(f)=(1/f)^{1/2}\end{eqnarray}
\begin{eqnarray}w_L(f)=(1-exp(-(f/0.5)^3))^{1/2}\end{eqnarray}
\begin{eqnarray}w_H(f)=(1+0.694S^2+0.241S^4+0.0557S^6+0.009664S^8+0.00134S^{10}+0.000155S^{12})^{-1/2}\end{eqnarray}

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\begin{eqnarray}v(t)=\sqrt[]{a_X^2(t)+a_Y^2(t)+a_Z^2(t)}\end{eqnarray}

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\begin{eqnarray}SI(h)=\int_{0.1}^{2.5}PSV(h,T)\cdot dT\end{eqnarray}

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\begin{eqnarray}a_{rms}=\sqrt[]{\frac{1}{T_d}\int_0^{T_d}[a(t)]^2\cdot dt\end{eqnarray}

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\begin{eqnarray}I_a=\frac{\pi}{2g}\int_0^{\infty}[a(t)]^2\cdot dt\end{eqnarray}

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Last-modified: 2018-06-25 (·î) 10:03:25 (2126d)