: the circle with the diameter $\xi o$; therefore the circle in the cylinder whose % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig04.png} \begin{center} {\small Fig.~4.} \end{center} \end{wrapfigure} % diameter is $\mu\nu$ is in its present position, in equilibrium at the point $\alpha$ with the circle whose diameter is $\xi o$ if this be transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity. And the center of gravity of the circle whose diameter is $\mu\nu$ is at $\sigma$, that of the circle whose diameter is $\xi o$ when its position is changed, is $\theta$, and we have the inverse proportion, $\theta\alpha : \alpha\sigma$ = the circle with the diameter $\mu\nu$ : the circle with the diameter $\xi o$. In the same way it can be shown that if another straight line be drawn in the parallelogram $\epsilon\gamma \| \beta\gamma$ the circle formed in the cylinder, will in its present position be in equilibrium at the point $\alpha$ with that formed in the segment of the right conoid if the latter is so transferred to $\theta$ on the scale-beam that $\theta$ is its center of gravity. Therefore if the cylinder and the segment of the right conoid are filled up then the cylinder in its present position will be in equilibrium at the point $\alpha$ with the segment of the right conoid if the latter is transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity. And since these magnitudes are in equilibrium at $\alpha$, and $\kappa$ is the center of gravity of the cylinder, if $\alpha\delta$ is bisected at $\kappa$ and $\theta$ is the center of gravity of the segment transferred to that point, then we have the inverse proportion $\theta\alpha : \alpha\kappa$ = cylinder : segment. But $\theta\alpha = 2\alpha\kappa$ and also the cylinder = 2 $\times$ segment. But the same cylinder is 3 times as great as the cone whose base is the circle on the diameter $\beta\gamma$ and whose vertex is at $\alpha$; therefore it is clear that the segment is one and one half times as great as the same cone. \section*{Proposition V} That the center of gravity of a segment of a right conoid which is cut off by a plane perpendicular to the axis, lies on the straight line which is the axis of the segment divided in such a way that the portion at the vertex is twice as great as the remainder, may be perceived by our method in the following way: