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:

Let a segment of a right conoid cut off by a plane perpendicular to the axis be cut by another plane through the axis, and let the intersection in its surface be the parabola αβγ [Fig. 5] and let the common line of intersection of the plane which cut off the segment and of the intersecting plane be βγ; let the axis of the segment and the diameter of the parabola αβγ be αδ; produce δα so that αθ = αδ and imagine δθ to be a scale-beam with its center at α; then inscribe a cone in the segment with the lateral boundaries βα and αγ and in the parabola draw a straight line ξo βγ and let it cut the parabola in ξ and o and the lateral boundaries of the cone in π and ρ. Now because ξσ and βδ are drawn perpendicular to the diameter of the parabola, δα: ασ = βδ²: ξσ² [Quadr. parab. 3]. But δα: ασ = βδ: πσ = βδ²: βδ × πσ, therefore also βδ²: ξσ² = βδ²: βδ × πσ. Consequently ξσ² = βδ × πσ and βδ: ξσ = ξσ: πσ, therefore βδ: πσ = ξσ²: σπ². But βδ: πσ = δα: ασ = θα: ασ, therefore also θα: ασ = ξσ²: σπ². On ξo con Fig. 5. struct a plane perpendicular to αδ; this will intersect the segment of the right conoid in a circle whose diameter is ξo and the cone in a circle whose diameter is πρ. Now because θα: ασ = ξσ²: σπ² and ξσ²: σπ² = the circle with the diameter ξo: the circle with the diameter πρ, therefore θα: ασ = the circle whose diameter is ξo: the circle whose diameter is πρ. Therefore the circle whose diameter is ξo will in its present position be in equilibrium at the point α with the circle whose diameter is πρ when this is so transferred to θ on the scale-beam that θ is its center of gravity. Now since σ is the center of gravity of the circle whose diameter is ξo in its present position, and θ is the center of gravity of the circle whose diameter is πρ if its position is changed as we have said, and inversely θα: ασ = the circle with the diameter ξo: the circle with the diameter πρ, then the circles are in equilibrium at the point α. In the same way it can be shown that if another straight line is drawn in the parabola βγ and on this line last drawn a plane is constructed perpendicular to αδ, the circle formed in the segment of the right conoid will in its present position be in equilibrium at the point α with the circle formed in the cone, if the latter is transferred and so arranged on the scale-beam at θ that θ 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 α with all circles of the cone if the latter are transferred and so arranged on the scale-beam at the point θ that θ is their center of gravity. Therefore also the segment of the right conoid in its present position will be in equilibrium at the point α with the cone if it is transferred and so arranged on the scale-beam at θ that θ is its center of gravity. Now because the center of gravity of both magnitudes taken together is α, but that of the cone alone when its position is changed is θ, then the center of gravity of the remaining magnitude lies on αθ extended towards α if ακ is cut off in such a way that αθ: ακ = segment: cone. But the segment is one and one half the size of the cone, consequently αθ = 3/2 ακ and κ, the center of gravity of the right conoid, so divides αδ that the portion at the vertex of the segment is twice as large as the remainder.