and substituting for ${\displaystyle \mathrm {H} }$ its value ${\displaystyle {\frac {V^{2}}{2g}}}$

${\displaystyle V=g\times {\frac {aa^{\prime }(a^{\prime }-a)}{2\cos ^{2}{e}(ab^{\prime }-a^{\prime }b)}}}$

now we have for the "camine"

${\displaystyle a={\text{ 7.4 feet, at the final point of repose}}}$

${\displaystyle b={\text{ 4.33 feet}}}$

and for the south end wall

${\displaystyle a^{\prime }={\text{ 14 feet, }}b^{\prime }={\text{ 30 feet.}}}$

Therefore

${\displaystyle -4.33=7.4\tan {e}-{\frac {13.69}{\mathrm {H} \cos ^{2}{e}}}}$

${\displaystyle -30=14\tan {e}-{\frac {49}{\mathrm {H} \cos ^{2}{e}}}}$

solving we find

${\displaystyle \mathrm {Tan\,} {e}=1.357{\text{ and }}e=53^{\circ }37^{\prime }}$

and horizontal velocity ${\displaystyle V_{,}=}$ 8.03 feet per second and

${\displaystyle \mathrm {V} =V_{,}\sec {e}=13.490{\text{ feet per second.}}}$

The difference between the former calculation and this is = 0.627 feet per second, or little more than half a foot per second. If we take the mean of the two determinations we find

${\displaystyle V=13.176{\text{ feet per second.}}}$

There can be little doubt, that the small bit of brickwork imposed over the lintel block, of this "camine," in moving at first, along with the lintel, and with a longer radius at the commencement of the trajectory, communicated to the lintel block, a velocity slightly greater, than alone it would have been projected with, and that hence the velocity deduced from the measurements of the axes of ${\displaystyle a}$ and ${\displaystyle b}$, may be slightly in excess. The true maximum velocity