This theory is based on the principle of minimum energy. According to this principle, angle will take a value such that total work done in cutting is a minimum.

As the work done depends only on the cutting force component Fp, then angle should assume such a value as to make Fp should be minimum for given.

From merchants circle diagram,

$F_{p}=R \cdot \cos (\beta-\gamma)$

$F_{p}=\frac{F_{s} \cdot \cos (\beta-\gamma)}{\cos \cdot \cot (\beta-\gamma)}$

$F_{p}=\frac{z \cdot b \cdot t \cdot \cos (\beta-\gamma)}{\sin \phi \cdot \cos (\beta-\gamma)}$

Where $Z =$ mean shear stress

$B =$ width of the chip

$t =$ uncut chip thickness.

If $\tau'$the shear strength of the material being cut, shear will occur when the shear strength on the plane defined by angle $\theta$ be completely equal to the shear strength $\tau'$ of the materials.

For a simple analytic it is assumed to be constant.

$F_{p}=\frac{\tau^{\prime} \cdot b t \cdot \cos (\beta-\gamma)}{\sin \emptyset \cdot \cos (\emptyset+\beta-\gamma)}$

For minimum $Fp$,

$\frac{d F_{p}}{d \emptyset}=0$

$\therefore \tau^{\prime} b t \cdot \cos (\beta-\gamma)[\cos \emptyset \cdot \cos (\emptyset+\beta-\gamma)-\sin \emptyset . \sin (\emptyset+\beta-\gamma)]=0$

$\therefore \cos (2 \emptyset+\beta-\gamma)=0$

Which gives a minimum value, or,

$2 \emptyset+\beta-\gamma=\frac{7}{3}$ Or, $\emptyset=\frac{\pi}{4}+\frac{\gamma}{2}-\frac{\beta}{2}$