= Optimal normal-fault dip relation
{title2=$\delta=45^\circ+\varphi/2$}
For a cohesionless, optimally oriented <normal fault> with vertical maximum compression and horizontal minimum compression,
$$
\delta=45^\circ+\frac\varphi2,\qquad\mu_f=\tan\varphi=\tan(2\delta-90^\circ),
$$
where $\delta$ is <fault dip> and $\varphi$ the <internal friction angle>. This is a geometric inference under the model, not a measurement of friction from dip alone.
For a compression-positive <Mohr circle>, put $C=(\sigma_1'+\sigma_3')/2$ and $R=(\sigma_1'-\sigma_3')/2$. Tangency to the cohesionless <Mohr-Coulomb failure criterion> $\tau=\mu_f\sigma_n'$ gives $R/C=\sin\varphi$. The contact point is $(\sigma_n',\tau)=(C\cos^2\varphi,C\sin\varphi\cos\varphi)$. The physical plane-normal angle to $\sigma_1'$ is half the corresponding central angle, giving the stated dip. Equivalently the plane itself makes $45^\circ-\varphi/2$ with $\sigma_1'$.
The associated stress ratio is $\sigma_1'/\sigma_3'=(1+\sin\varphi)/(1-\sin\varphi)$. A reactivated nonoptimal plane, nonzero <cohesion> or unknown <principal stress> directions defeats the simple dip-only inversion.
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