therefore

${\displaystyle l={\frac {k^{2}}{\alpha }}={\frac {2{\sqrt {2}}\alpha }{3}}}$

and

${\displaystyle \cos {\phi }={\frac {1}{\sqrt {2}}}}$

substituting these values in (II.) we find

${\displaystyle V^{2}={\frac {8}{3}}g\alpha ({\sqrt {2}}-1)}$

(III.)

and the following geometrical construction is obtained from this expression.

Let ${\displaystyle \delta =}$ the difference between the diameter and side of the cube ${\displaystyle =\alpha \left({\sqrt {2}}-1\right)}$

${\displaystyle V^{2}=2g\times \left({\frac {4}{3}}\delta \right)}$

(IV.)

Or, the height due to the horizontal velocity of the wave of shock is equal to four thirds of the difference of the diameter and side of the cube.

2nd. In the case of a solid rectangular parallelopiped overturned (Fig. 95).

The altitude being ${\displaystyle =\alpha }$, side of the base ${\displaystyle =\beta }$, and ${\displaystyle tan{\phi }={\frac {\beta }{\alpha }}}$, we have

${\displaystyle k^{2}={\frac {\alpha ^{2}+\beta ^{2}}{3}}~~~~~~~~a={\frac {1}{2}}{\sqrt {\alpha ^{2}+\beta ^{2}}}}$