Optimal normal-fault dip relation ($\delta =45^{\circ }+\phi /2$)

For a cohesionless, optimally oriented normal fault with vertical maximum compression and horizontal minimum compression,where $\delta$ is fault dip and $\phi$ 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 $C=({\sigma }_1^{\prime }+{\sigma }_3^{\prime })/2$ and $R=({\sigma }_1^{\prime }-{\sigma }_3^{\prime })/2$. Tangency to the cohesionless Mohr-Coulomb failure criterion $\tau ={\mu }_f{\sigma }_n^{\prime }$ gives $R/C=\sin \phi$. The contact point is $({\sigma }_n^{\prime },\tau )=(C{\cos }^2\phi ,C\sin \phi \cos \phi )$. The physical plane-normal angle to ${\sigma }_1^{\prime }$ is half the corresponding central angle, giving the stated dip. Equivalently the plane itself makes $45^{\circ }-\phi /2$ with ${\sigma }_1^{\prime }$.

The associated stress ratio is ${\sigma }_1^{\prime }/{\sigma }_3^{\prime }=(1+\sin \phi )/(1-\sin \phi )$. A reactivated nonoptimal plane, nonzero cohesion or unknown principal stress directions defeats the simple dip-only inversion.