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Table 2 Reluctance calculation result

From: Analytical Modelling and Experiment of Novel Rotary Electro-Mechanical Converter with Negative Feedback Mechanism for 2D Valve

Reluctance

Result

\({R}_{\mathrm{a}}\)

\({R}_{\mathrm{a}1}={R}_{\mathrm{a}2}=\frac{3.846}{{\mu }_{0}{L}_{\mathrm{g}1}}\)

\({R}_{\mathrm{b}}\)

\({R}_{\mathrm{b}1}={R}_{\mathrm{b}2}=\frac{1.923}{{\mu }_{0}{L}_{\mathrm{g}1}}\)

\({R}_{\mathrm{b}}^{^{\prime}}\)

\({R}_{\mathrm{b}1}^{^{\prime}}={R}_{\mathrm{b}2}^{^{\prime}}=\frac{1.923}{{\mu }_{0}{L}_{\mathrm{g}2}}\)

\({R}_{\mathrm{c}}\)

\({R}_{\mathrm{c}1}=\frac{\pi }{{\mu }_{0}{L}_{\mathrm{g}1}\mathrm{ln}\left(1+\frac{2m}{g-x}\right)}\)

\({R}_{\mathrm{c}2}=\frac{\pi }{{\mu }_{0}{L}_{\mathrm{g}1}\mathrm{ln}\left(1+\frac{2m}{g+x}\right)}\)

\({R}_{\mathrm{d}}\)

\({R}_{\mathrm{d}1}=\frac{\pi }{2{\mu }_{0}{L}_{\mathrm{g}1}\mathrm{ln}\left(1+\frac{m}{g-x}\right)}\)

\({R}_{\mathrm{d}2}=\frac{\pi }{2{\mu }_{0}{L}_{\mathrm{g}1}\mathrm{ln}\left(1+\frac{m}{g+x}\right)}\)

\({R}_{\mathrm{d}}^{^{\prime}}\)

\({R}_{\mathrm{d}1}^{^{\prime}}=\frac{\pi }{2{\mu }_{0}{L}_{\mathrm{g}2}\mathrm{ln}\left(1+\frac{m}{g-x}\right)}\)

\({R}_{\mathrm{d}2}^{^{\prime}}=\frac{\pi }{2{\mu }_{0}{L}_{\mathrm{g}2}\mathrm{ln}\left(1+\frac{m}{g+x}\right)}\)

\({R}_{\mathrm{e}}\)

\({R}_{\mathrm{e}1}=\frac{12.987}{{\mu }_{0}\left(g-x\right)}, {R}_{\mathrm{e}2}=\frac{12.987}{{\mu }_{0}\left(g+x\right)}\)

\({R}_{\mathrm{f}}\)

\({R}_{\mathrm{f}1}={R}_{\mathrm{f}2}=\frac{4}{{\mu }_{0}m}\)

\({R}_{\mathrm{x}}\)

\({R}_{\mathrm{x}1}=\frac{g-x}{{\mu }_{0}{A}_{\mathrm{g}}}, {R}_{\mathrm{x}2}=\frac{g+x}{{\mu }_{0}{A}_{\mathrm{g}}}\)