Stray load losses are modeled dependent on square of current, but without scaling them to zero at no-load current. The losses do not cause a voltage drop in the electric circuit, they are dissipated through an equivalent braking torque at the shaft. The dependency of the stray load torque on the angular velocity is modeled by the exponent $p_{\mathrm{\omega }}$.

$\mathrm{\tau }\&equals;-{\mathrm{\tau }}_{\mathrm{ref}}{\left(\frac{i}{I_{\mathrm{ref}}}\right)}^2\{\begin{array}{cc}0& P_{\mathrm{ref}}<0\\ {\left(\frac{\mathrm{\omega }}{{\mathrm{\omega }}_{\mathrm{ref}}}\right)}^{p_{\mathrm{\omega }}}& 0\le \mathrm{\omega }\\ -{\left(-\frac{\mathrm{\omega }}{{\mathrm{\omega }}_{\mathrm{ref}}}\right)}^{p_{\mathrm{\omega }}}& \mathrm{otherwise}\end{array}$

$v\&equals;v_p-v_n\&equals;0$

$i\&equals;i_p\&equals;-i_n$

$\mathrm{\phi }\&equals;{\mathrm{\phi }}_{\mathrm{flange}}-{\mathrm{\phi }}_{\mathrm{support}}$

$\mathrm{\omega }\&equals;\stackrel{\&period;}{\mathrm{\phi }}$

$\mathrm{\tau }\&equals;{\mathrm{\tau }}_{\mathrm{support}}\&equals;-{\mathrm{\tau }}_{\mathrm{flange}}$

${\mathrm{\tau }}_{\mathrm{ref}}\&equals;\frac{P_{\mathrm{ref}}}{{\mathrm{\omega }}_{\mathrm{ref}}}$

$\mathrm{lossPower}\&equals;-\mathrm{\tau }\mathrm{\omega }$

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