\sigma\xi^2 = \sigma\pi^2 : \sigma\pi \times \pi\mu$, consequently $\mu\pi \times \pi\sigma = \sigma\xi^2$. If $\pi\sigma^2$ is added to both sides then $\mu\sigma \times \sigma\pi = \pi\sigma^2 + \sigma\xi^2$. Therefore $\theta\alpha : \alpha\sigma = \mu\sigma^2 : \pi\sigma^2 + \sigma\xi^2$. But $\mu\sigma^2 : \sigma\xi^2 + \sigma\pi^2 =$ the circle in the cylinder whose diameter is $\mu\nu :$ the circle with the diameter $\xi o$ + the circle with the diameter $\pi\rho$; hence the circle whose diameter is $\mu\nu$ will in its present position be in equilibrium at the point $\alpha$ with the two circles whose diameters are $\xi o$ and $\pi\rho$ when they are transferred and so arranged on the scale-beam at the point $\alpha$ that $\theta$ is the center of gravity of both; and $\theta$ is the center of gravity of the two circles combined whose diameters are $\xi o$ and $\pi\rho$ when their position is changed, % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig03.png} \begin{center} {\small Fig.~3.} \end{center} \end{wrapfigure} % hence $\theta\alpha : \alpha\sigma =$ the circle with the diameter $\mu\nu :$ the two circles whose diameters are $\xi o$ and $\pi\rho$. In the same way it can be shown that if another straight line is drawn in the parallelogram $\lambda\zeta \| \epsilon\zeta$ and on this line last drawn a plane is constructed perpendicular to $\alpha\gamma$, then likewise the circle produced in the cylinder will in its present position be in equilibrium at the point $\alpha$ with the two circles combined which have been produced in the spheroid and in the cone respectively when they are so transferred to the point $\theta$ on the scale-beam that $\theta$ is the center of gravity of both. Then if cylinder, spheroid and cone are filled with such circles, the cylinder in its present position will be in equilibrium at the point $\alpha$ with the spheroid $+$ the cone if they are transferred and so arranged on the scale-beam at the point $\alpha$ that $\theta$ is the center of gravity of both. Now $\kappa$ is the center of gravity of the cylinder, but $\theta$, as has been said, is the center of gravity of the spheroid and cone together. Therefore $\theta\alpha : \alpha\kappa =$ cylinder $:$ spheroid $+$ cone. But $\alpha\theta = 2\alpha\kappa$, hence also the cylinder = 2 $\times$ (spheroid $+$ cone) = 2 $\times$ spheroid + 2 $\times$ cone. But the cylinder = 3 $\times$ cone, hence 3 $\times$ cone = 2 $\times$ cone + 2 $\times$ spheroid. Subtract 2 $\times$ cone from both sides; then a cone whose axes form the triangle $\alpha\epsilon\zeta$ = 2 $\times$ spheroid. But the same cone = 8 cones whose axes form the $\Delta\alpha\beta\delta$; hence 8 such cones = 2 $\times$ spheroid, 4 $\times$ cone = spheroid; whence it follows that a spheroid is four times as great as a cone whose vertex is at $\alpha$, and whose base is the circle on the diameter $\beta\delta$ perpendicular to