straight line is drawn in the parabola $\| \beta\gamma$ and on this line last drawn a plane is constructed perpendicular to $\alpha\delta$, the circle formed in the segment of the right conoid will in its present position be in equilibrium at the point $\alpha$ with the circle formed in the cone, if the latter is transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity. Therefore if the segment and the cone are filled up with circles, all circles in the segment will be in their present positions in equilibrium at the point $\alpha$ with all circles of the cone if the latter are transferred and so arranged on the scale-beam at the point $\theta$ that $\theta$ is their center of gravity. Therefore also the segment of the right conoid in its present position will be in equilibrium at the point $\alpha$ with the cone if it is transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity. Now because the center of gravity of both magnitudes taken together is $\alpha$, but that of the cone alone when its position is changed is $\theta$, then the center of gravity of the remaining magnitude lies on $\alpha\theta$ extended towards $\alpha$ if $\alpha\kappa$ is cut off in such a way that $\alpha\theta : \alpha\kappa$ = segment : cone. But the segment is one and one half the size of the cone, consequently $\alpha\theta = \frac{3}{2}\alpha\kappa$ and $\kappa$, the center of gravity of the right conoid, so divides $\alpha\delta$ that the portion at the vertex of the segment is twice as large as the remainder. \section*{Proposition VI} [The center of gravity of a hemisphere is so divided on its axis] that the portion near the surface of the hemisphere is in the ratio of $5 : 3$ to the remaining portion.