= Gear effective face width
{title2=$b_{\rm eff}$}
<Gear effective face width> is the usable axial width participating in a stated cylindrical gear mesh. It must be distinguished from each component's nominal <gear face width>, and from a claim that every part of that width carries equal load. The normal/transverse tooth geometry and <gear overlap ratio> use this engaged width, while <gear face load distribution> describes how the load actually occupies it.<Elements of Metric Gear Technology by SDP-SI>[D815 technical section, §11, Table11-3, axial overlap ratio; §17.2.12, tooth-flank load distribution]{ref}
An elementary geometric model treats the two toothed widths as intervals $I_1=[a_1,b_1]$ and $I_2=[a_2,b_2]$ on their common axial coordinate. Their available overlap is
$$
b_{\rm eff}=\max\{0,\min(b_1,b_2)-\max(a_1,a_2)\}.
$$
For one20mm-wide gear occupying $[-10,10]\,\mathrm{mm}$ and a16mm-wide mate centered at $6\,\mathrm{mm}$, the second interval is $[-2,14]\,\mathrm{mm}$. Their overlap is only12mm. The mean line load for a hypothetical uniformly shared normal <force> of1200N is then100N/mm, rather than the60N/mm obtained by using the first gear's entire nominal width.
With <gear normal module> $m_n=2\,\mathrm{mm}$ and <gear helix angle> $|\beta|=30\,{}^{\circ}$, <gear axial pitch> is $p_x=4\pi\,\mathrm{mm}$. The12mm overlap gives geometric <gear overlap ratio> $b_{\rm eff}/p_x=3/\pi\approx0.9549$, compared with $5/\pi\approx1.5915$ for20mm. Axial positioning can therefore alter overlap without changing the ideal <gear ratio>.
The interval calculation assumes the flanks remain usable throughout their common extent. Relief, damaged regions or partial contact can require a more specific working-width convention. Uneven loading is not generally corrected by shrinking width until a uniform-load formula looks plausible; a stated load-distribution model is needed to predict local loading.
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