ARTICLE III. A Demonstration that the Ellipse, when viewed in a certain Position, appears circular. By S. SIR, (To Dr. Thomson.) Jan. 8, 1816. It is well known that the circle, if looked upon obliquely, will be projected into an ellipse; but I am not aware that the converse of this proposition has been demonstrated by showing that an ellipse, if viewed in a certain position, will appear circular. This has been established in the following theorems; but it was not the primary object with which they were drawn up. They were occasioned by the wish of getting a scalene cone turned truly in a lathe. Many good workmen assured me that it was impossible; and all the cones of this kind which I have seen (with the exception of that which I am about to mention) have been cut by hand. In thinking on the subject, it appeared that the best way would be to get a cone turned in the first instance with an elliptical base; but here again I met with a difficulty which was not surmounted till within these few months, by a very ingenious friend, who devised the means of executing exactly what I wished. By a mechanical method of trial, he afterwards found where he might cut this elliptical cone obliquely, so that the base should become a circle; but it seemed more satisfactory to investigate the problem mathematically, and I here send you the result. It would be unjust, however, not to add that the first lemma is taken from Dr. Robertson's Conic Sections (8vo. Oxford, 1802), and that several parts of the proposition were suggested by a recollection of the methods used in the first book of that valuable treatise. S. Lemma 1.-If from any point, I, of a straight line, NB (Plate XLV.* Fig. 1), a perpendicular, I L, be drawn; and if in all cases the rectangle under NI, I B, shall be equal to the square of IL, the curve passing through N LB shall be a semicircle. For bisect N B in C, and join C L. Then (5, ii.) N I, IB, + ICC B2; therefore by hypothesis I L + ICC B2; but (47, i.) I L2 + ICC L2; consequently C B2 = C L'. The same would hold wherever the point is taken; therefore the locus of the points L must be a curve, which would be generated by the extremity, L, of the given straight line, C L, when its other extremity, C, is fixed, and the straight line revolves about it. Lemma 2.-Let VAB (Fig. 2) be an isosceles triangle, and OG a straight line longer than the base; then if VT be taken, such that O G A B2: O G2 :: V B2 : V T2, we shall have A B2 : For A B OG :: VT - V B: OG: AT, TB: VT. : VT; and when VC is drawn perpendicular to (and therefore bisecting) the base, V T2 V B2 = (47, i.) V C2 + C T2 - V C2 - CBCTC BAT, TB (6, ii.). *The lower division of the Plate. Definition. The straight line joining the vertex of the cone and the centre of the ellipse (which forms the base) is the axis of the cone. The only case which bears upon the principal object of the present investigation is when this axis is perpendicular to the base; to this case, therefore, the first, second, and third, propositions are confined; the second, indeed, would be generally true, whatever inclination were given to the axis; but it is not my intention at present to go into the full consideration of the sections of the elliptic cone. Prop. 1. Any plane passing through the axis will be perpendicular to the base, and its common section with the conical superficies will be an isosceles triangle. To The first part of the proposition will be true by Eucl. (18, xi.) prove the second part, let V C (Fig. 3) be the axis of the elliptical cone V A C B, and C the centre of the elliptical base. Then it may be shown exactly, as in the circular cone, that the common sections of the plane and conical superficies will be the straight lines VA, V B; consequently V A B is a triangle. Now as A B is a diameter of the elliptical base, and C is the centre, A C = CB. Hence A C2 + V C2 C B + V C2; but by (47, i.) A C2 + VC2 V A2, and C B2 + V CV B2; therefore V A = V B.-Q. E. D. Prop. 2.-If the conical superficies be cut by a plane parallel to the base, the common section will be an ellipse similar to the base. Let (in Fig. 4) DHEF be the common intersection of the conical superficies, and of the plane parallel to the base; then DH E is an ellipse similar to the base A G B. = Let a plane pass through the axis, and likewise through any diameter, A B, of the base. Let its common section with the plane D H E be DE, and let the axis meet the plane D H E F in F. Then (16, xi.) D E is parallel to A B, and DT: AC:: VF: VC FE:C B. Hence DF: AC:: FE: C B, and A C = C B; therefore D F FE. Now this will be true, whatever may be the position of A B and D E; therefore any line terminated by D H E, and passing through F, will be bisected in that point. Again, let G C be a semidiameter conjugate to A C, and through GC, VC pass a plane, and let its common section with the plane DEHF be HF. Then, as before, H F is parallel to G C, and DF:AC: HF: CG; therefore alternately DF: HF:: AC: CG, and (10, xi.) the angle D F H is equal to the angle ACG; and as this will be true of any conjugate diameters, it follows that DHE is an ellipse similar to A G B, and that its centre is at F. Cor D F F H2 :: A C2: C G. Hence the rectangle under the abscissæ of D E : the square of its ordinate :: A C2 : C G2. Prop. 3.-Let V ABG be a cone with an elliptic base, of which GO is the greater, and A B the smaller, axis; and let the common section of a plane passing through the axis of the cone and the minor axis, A B, of the ellipse, be the isosceles triangle V A B: In A B produced take the point T, such that O G A B2: O Go :: VB VT. Then by lemma 2, A B2: O G2 :: AT, TB: V T% 2 Through B in the triangle V A B draw B N parallel to V T, and through BN pass a plane perpendicular to the plane V A B, and let the common section of this plane with conical superficies be the line N LBK, N LBK shall be a circle. 1 In BN take any point I, and through I pass a plane parallel to the base. Let the common intersection of this plane, and of the conical superficies, be D KEL, which by Prop. 2 will be an ellipse similar to the base. Let the common intersection of the planes DLE K and NLB K be L K. Now as the axis of the cone is perpendicular to the base, the plane A G BO (18, xi.) is at right angles to the plane VA B, consequently D L KE is at right angles to V AB; but the plane NLBK was made at right angles to VA B; therefore (19, xi.) LK is at right angles to the plane VAB and LIN, LID are both right angles. 1. Now as B N is parallel to V T, and D E is parallel to A T, the triangles DIN, A T V, and the triangles IE B, VBT, are similar. Hence DI: NI:: AT: VT, and IE: IB:: TB:VT. Therefore D I, IE: NI, IB:: AT, TB:V T2; but by construction A T, TB: VT:: A B2: O G. Therefore DI, IE: NJ, IB: A B2: O G2 :: A C2: C G2. Lastly, by Prop. 2, D E is the minor axis of DK EL, and by the corollary DI,IE: IL: A C: C G. But it has been shown that DI, IE: NI, IB:: A C2: C G2; therefore D I, IE: I L2 :: DI, IE: NI, IB; consequently I L NI, IB, and by lemma 1, NLB is a semicircle. = Cor. 1.-As√OG-AB:OG:: VB: VT,OG: OG-AB :: sine of VA B: sine of VTA or NBA. If the cone becomes a cylinder, V A B becomes a right angle, and OG: √OG — A B2 :: radius: sine of N B A. Cor. 2.-V NL BK will become a scalene cone; therefore any plane parallel to NLB K, or any plane which forms the subcontrary section, will give a circle. ARTICLE IV. Letter of Dr. Schübler, Professor of Natural Philosophy and Che mical Agriculture in the Institute, of Hofwyl, to Professor M. A. Pictet, on the Physical Analysis of Soils.* Hofwyl, July 14, 1815. In consequence of your desire, during your recent visit to Hofwyl, to know the result of my researches on the physical qualities of arable soils, I do myself the honour of transmitting to you the following comparative table. I shall likewise explain the way in which my experiments were performed: * Translated from the Bibliotheque Britannique, Agriculture, p. 248, July, 1815, |