Let $\alpha\beta\gamma\delta$ [Fig.~2] be the largest circle of a sphere and $\alpha\gamma$ and $\beta\delta$ its diameters perpendicular to each other; let there be in the sphere a circle on the diameter $\beta\delta$ perpendicular to the circle $\alpha\beta\gamma\delta$, and on this perpendicular circle let there be a cone erected with its vertex at $\alpha$; producing the convex surface of the cone, let it be cut through $\gamma$ by a plane parallel to its base; the result will be the circle perpendicular to $\alpha\gamma$ whose diameter will be $\epsilon\zeta$. On this circle erect a cylinder whose axis $= \alpha\gamma$ and whose vertical boundaries are $\epsilon\lambda$ and $\zeta\eta$. Produce $\gamma\alpha$ making $\alpha\theta = \gamma\alpha$ and think of $\gamma\theta$ as a scale-beam with its center at $\alpha$. Then let $\mu\nu$ be any straight line whatever drawn $\| \beta\delta$ intersecting the circle $\alpha\beta\gamma\delta$ in $\xi$ and $o$, the diameter $\alpha\gamma$ in $\sigma$, the straight line $\alpha\epsilon$ in $\pi$ and $\alpha\zeta$ in $\rho$, and on the straight line $\mu\nu$ construct a plane perpendicular to $\alpha\gamma$; it will intersect the cylinder in a circle on the diameter $\mu\nu$; the sphere $\alpha\beta\gamma\delta$, in a circle on the diameter $\xi o$; the cone $\alpha\epsilon\zeta$ in a circle on the diameter $\pi\rho$. Now because $\gamma\alpha \times \alpha\sigma = \mu\sigma \times \sigma\pi$ ( for $\alpha\gamma = \sigma\mu$, $\alpha\sigma = \pi\sigma$), and $\gamma\alpha \times \alpha\sigma = \alpha\xi^2 = \xi\sigma^2 + \alpha\pi^2$ then $\mu\sigma \times \sigma\pi = \xi\sigma^2 + \sigma\pi^2$. Moreover, because $\gamma\alpha : \alpha\sigma = \mu\sigma : \sigma\pi$ and $\gamma\alpha = \alpha\theta$, therefore $\theta\alpha : \alpha\sigma = \mu\sigma : \sigma\pi = \mu\sigma^2 : \mu\sigma \times \sigma\pi$. But it has been proved that $\xi\sigma^2 + \sigma\pi^2 = \mu\sigma \times \sigma\pi$; hence $\alpha\theta : \alpha\sigma = \mu\sigma^2 : \xi\sigma^2 + \sigma\pi^2$. But it is true that $\mu\sigma^2 : \xi\sigma^2 + \sigma\pi^2 = \mu\nu^2 : \xi\alpha^2 + \pi\rho^2 =$ the circle in the cylinder whose diameter is $\mu\nu :$ the circle in the cone whose diameter is $\pi\rho$ + the circle in the sphere whose diameter is $\xi o$; hence $\theta\alpha : \alpha\sigma =$ the circle in the cylinder $:$ the circle in the sphere $+$ the circle in the cone. Therefore the circle in the cylinder in its present position will be in equilibrium at the point $\alpha$ with the two circles whose diameters are $\xi o$ and $\pi\rho$, if they are so transferred to $\theta$ that $\theta$ is the center of gravity of both. In the same way it can be shown that when another straight line is drawn in the parallelogram $\xi\lambda \| \epsilon\zeta$, and upon it a plane is erected perpendicular to $\alpha\gamma$, the circle produced in the cylinder in its present position will be in