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p2q=kr.

Let the cubic be thus denoted:

The critical value (k=5·1458) determines the bitangential cubic. The cubic is bipartite or unipartite, according as «><5·1458; equiharmonic if x=1·67 or -·17; harmonic if x= -15.

There may be two real asymptotes, since there is one bitangent in this division. II. When the satellite-point is on the bounding sextic.

In this case there are two critical values of the parameter, corresponding to the bitangential and inflexional cubics, viz. 11·09 and 08. The curve is bipartite if K>11.09.

III. When the satellite is inside the quartic, and is (1) not collinear, and (2) is collinear with the foci.

(1) There are three bitangents when k=47, 07, 06. For higher values of K, the curve is bipartite; between 47 and 07, unipartite; then bipartite and below the bitangential curve unipartite.

There may be four asymptotes.

(2) Let the cubic be thus denoted:

λ(1+bέ)+(1+c§)(§2+n2)=0.

There are two bitangents with real and imaginary contact, as is thus shown:— 4+λ(27c2-18bc-b2)+4x2b3c=0.

For the inflexional genus, the discriminant gives the condition

646'c (27c2-18bc-b2)2.

This may be resolved into two factors:

(b−c) (b—9c)3.

The first factor resolves the cubic into a point and a circle; the second factor indicates the cissoid:

(3+b§)3+b2(9+b)n2=0.

The satellite-point in this case is the apse in the quartic bounding curve.

IV. If the satellite-point be at infinity (1) not collinear, (2) collinear, with the two foci.

(1) There are two bitangential or a single inflexional, or no bitangential form, according as the satellite lies within, upon, or beyond the quartic curve. One asymptote connects the double focus with the satellite; the other three concur in the point (=); the polar conic of the line at infinity degenerates into these points.

(2) There may be two asymptotes, which unite in a bitangent, for a special value of the parameter.

V. If the double focus is at infinity, (1) not collinear, (2) collinear with the single focus and satellite-point.

The cubic has in all cases a bitangent; and for a particular value of the parameter two bitangents coincide in a stationary tangent at a point of inflexion. The inflexional cubic in (2) is the semicubical parabola.

VI. If the single focus is at infinity (1) not collinear, (2) collinear with the double focus and satellite.

There are two bitangential forms, but no inflexional case. The reciprocals in (2) are Newton's defective hyperbole, with diameters and double foci.

VII. If the single focus and satellite-point are both at infinity.

The curve is central and parabolic, with a cusp at infinity, but cannot have a bitangent. Its single asymptote connects the double focus with the satellite. All cubics are equiharmonic of the form

λ+(√3§+n) (§2+n2).

It is thus denoted in Boothian coordinates:

λ+(a+bn) (§2+n2)=0.

The reciprocal is Newton's central species (38).

VIII. If the double focus and satellite-point are both at infinity.

There is an inflexional form in all cases, as appears from the equation to the system:

x2+(a+bn) (§2+n2)=0.

The reciprocal is a cusped cubic.

IX. If the double and single focus are both at infinity.

The line at infinity is an acu-bitangent in all cases, as is shown by the equation to the system :—

λ§2 (a§+bn)+(§2+n2)=0.

6. To find the asymptotes of this group of class-cubics.

Since the polar-point of an asymptote (p, q, r) lies on it, and also is at an infinite distance, the coordinates of an asymptote must satisfy two equations :

=λp2q+4A2r=0,.

and that to the polar conic of the line at infinity

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+ +
dp dy dr

λp(2q+p)+4A2=0,

one of whose foci, as we should anticipate, is the double focus.

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The four asymptotes touch a conic, whose foci are p=0, 2q+p=0. Hence also

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The elimination of r from (2) and (3) is a quartic equation. Hence there cannot be more than four asymptotes. Its discriminant is a factor of the discriminant of the ternary cubic (1). Two imaginary asymptotes always connect the double focus with the circular points at infinity.

If the satellite (which is always on the curve) be at infinity, its connector with the double focus is an asymptote. The extremities of two asymptotes may coincide in a bitangent.

7. The centre or polar point of the line at infinity is on AB, the connector of the foci, at the distance AB from A, the double focus. AC touches the cubic in C; BC touches it where it meets the line (B-ky=0).

On Spherical Class-cubics with Double Foci and Double Cyclic Arcs.
By HENRY M. JEFFERY, M.A.

1. This group may be denoted by line coordinates, as in plano :

where

kp2q+(6V)2r=0,

(6V)2=Σ(a2p2—2bcgr cos A),

and the coordinates p, q, r denote the sines of the perpendiculars from the vertices ABC on a tangent arc; and generally the symbols may denote the sines of arcs. There are four critic values of the parameter and four bitangential values. By partial differentiation, for a critic value,

2xpq+2raP=0: «p2+2rbQ=0: (6V)2+2rcR=0,

where Pap-bq cos C-cr cos B; and similar expressions denote Q, R; the linecoordinates of a tangent are referred to the polar triangle as one of reference. These conditions for a bitangent may be thus written :

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apP=(6V)2: 2bqQ=(6V)2 : 2crR=—(6V)2.

These equations denote three spherical ellipses, whose foci are in the several cases the points of reference A, B, C, and the corresponding points of reference of the polar triangle of ABC.

2. There is a sextic bounding curve, the locus of the satellite-point, when there is a point of inflexion; this curve is bipartite with an oval. If the satellite is within the oval, four critical values of the parameter yield bitangential forms of the cubic; if the satellite is on the oval, one inflexional and two bitangential cubies; if the satellite is outside the oval, two bitangential cubics. Its equation, in Güdermann's coordinates, is

{(4b2+3)y2+(b+x)(b+3x)}3=27(b2+1)(x2+y2){(x+b)2+y2}2.

If the two foci are a quadrant apart, b=x, and the bounding sextic becomes

4y2+1=3(x+y2)3.

3. All cubics with double foci have double cyclic arcs. Let the line-equation to these cubics be written

3ka2bp2q+3crƐ(a2p2 — 2bcrq cos A)=0.

Its equivalent point-equation may be thus arranged:—

(32+2ẞy cos A+ y2) { −12ẞy3×3 +3k2(8822 — a2y2+27a232+18a2ßy cos A 20aßy2cos C+36aß2y cos B)+12x[—ß3y - aß3 cos B+a1cos A-a3y cos C −3aß3y cos C+ ( − cos B+2 cos A cos C)a3 + (cos A-2 cos B cos C) a2ß2

—-(1+2 cos2 C)a2ßy]} +12(6V)2 (sin A) * {μ3‚2x2+2× (−ß3y — 4aß3 cos B

+4a2ß2 cos A+3a2ßy+2aß3y cos C)+(a2+2aß cos C+32)2} .

But, if V denote the volume of the tetrahedron constituted by the centre of the sphere and the angular points of the triangle of reference,

(GV)2=(a*a2+2bcẞy cos a)

=(aa+bẞ cos c+cy cos b)2+b2c2(82+28y cos A+y3).

Hence if p=0 be a double focus, its quadrantal polar (aa+bẞ cos c-cy cos b=0) is a double cyclic arc. (See §7.)

The proposition seems to be susceptible of simple proof and of generalization. 4. If a spherical curve have a multiple cyclic arc, it has at least a double focus. Let the triangle of reference be trirectangular (which assumption does not affect the generality of the proof), and let the quartic exhibit AB as a multiple cyclic

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In this form the imaginary lines (x+yi=0) are seen to meet AB in two coincident points I, J; the tangents at these points are these arcs CI, CJ: their point of concurrence is therefore a double focus. This proof seems applicable only to the case where the focus is the quadrantal pole of the cyclic arc, the points I, J being in this case the shadows of the circular points at infinity.

The argument may be also thus stated. The two lines

(x ±yi=0)

are common tangents to the curve, and to the imaginary sphere (x2+y2+z2=0) at their point of contact of a high order; their intersection is consequently a quadruple, or, if real values alone are considered, a double focus, and might be a multiple focus of a higher order.

5. On equiharmonic or neutral cubics with double foci. The cusps are collinear, and in this case S, an invariant of the cubic equation, =0.

(4 cos2 A-3)2+2(cos A+3 cos B cos C+2 cos A cos2 C)k+sin1 C=0.

There are two possible or coincident or impossible cases, according to the value of the parameter.

or

The two values coincide if

sin' C(4 cos" A-3)=(cos A+3 cos B cos C+2 cos A cos2 C)2,

(cot B+3 cos a cot C) (cot B+cos a cot C) +3 sin2 a=0,

the biangular equation to a conic.

That is, the locus of the double focus when the cusps are collinear, or the bounding curve, on either side of which equiharmonic values are or are not possible, is a spherical ellipse, whose cyclic arcs, real or imaginary, are perpendicular to the line connecting the single focus and satellite.

In plano the bounding line is thus denoted :-

a cot B+3 cot C=0, or cos C=- 26

The double focus is in a line, which cuts orthogonally the connector of the single focus and the satellite.

6. On harmonic cubics with double foci.

The invariant T=0.

coз A (9—8 cos2 A)x2+3 {6—4 cos2 A-9 cos2 B+3 cos2 C – 8 cos2 A cos2 C

- 12 cos Acos BcosC}x2+3sin3 C (cos A+3cos Bcos C+2 cos A cos2 C)×+ sin® C=0· For every position of the foci and satellite there is at least one value of the parameter which yields a harmonic cubic.

7. On the discriminant of the cubic.

Equate to zero in the point-equation (§ 3); besides the point of contact (B-ky=0), three tangential points are determined by the aggregate :

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ky (82+2ẞy cos A+y)-B(B2 sin2 B+23y sin B sin C cos a+y2 sin2 C).

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Since the anharmonic ratio of the lines connecting the tangential points depends upon the function 64-- the discriminant of the ternary cubic is simply found from this binary cubic :

{9K sin2 B+ (2x cos A-sin2 C) (k-2 sin B sin C cos a)}* +4{3 sin2 B(2k cos A-sin C)+(x-2 sin B sin C cos u)"} x{3K(K-2 sin B sin C cos a)-(2x cos A-sin2 C)}=0.

8. By dualising, this investigation is equally applicable to order-cubics with double cyclic arcs and double foci.

Résumé of Researches on the Inverse Problems of Moments of Inertia and of Moments of Resistance. By Professor GIUSEPPE JUNG (Milan).

In the study of the resistance of materials and the stability of constructions, the two following problems continually present themselves :—

I. To construct a plane figure (for example, the cross-section of a cylinder loaded in a given manner) of which we may suppose given the orientation, the form, the centre of gravity, and also the moment of inertia with respect to a given neutral axis.

II. To construct a plane section, given the orientation, the form, the centre of gravity, and also the moment of resistance* with respect to a given neutral axis.

* If e is an element of a plane section F, and y be the distance of the barycentre of e from an axis a measured in the direction A (A being a straight line making any angle with the axis x), then the moment of inertia of F with respect to a in the direction A is =ey=J, the Σ extending over the contour of F.

If, besides, v is the distance (measured parallel to A) of the barycentre O of F from the tangent to F parallel to r and furthest removed from O, then the moment of resistance Ꭻ of F with respect to a barycentric axis x, in the direction A, is defined to be the ratio In fact, if we multiply this ratio by a certain constant we have the ordinary moment of resistance of the section F.

These two problems have recently engaged my attention.

It is well known that engineers resolve these questions by tentative methods which sometimes require long calculations, and which besides are incapable of performance when the section is quite irregular; and this is why it is necessary to fix types (such as a Zorès iron, T's, I's, &c.) which, being decomposable into parts whose moments of inertia or moments of resistance can be determined analytically, are calculable.

By the simple and uniform graphical method which I have proposed, we can treat in the same manner as the simple sections (triangles, rectangles, &c.) the most complicated forms (such as a Zorès iron or even figures with arbitrary and irregular contours, whose equations would not admit of expression); so that, in order to render possible the solution of these important problems, we need sacrifice nothing, either from the economical or the aesthetic point of view.

I. Let F be the unknown figure that we wish to construct, J its moment of inertia in a direction A with respect to a given barycentric axis r*, F' a figure homothetical to F, O its centre of gravity, and x' a straight line parallel to a and passing through O'.

Let, besides, k be the (unknown) radius of gyration of F, in the direction A with respect to x, so that J=k2F; and let J' and ' be analogous quantities to J and k, relating to F'.

Finally, let us suppose two orthogonal axes u, w drawn anywhere, on which we have respectively the segments UA=WA=1, A being the point of intersection of u and w.

Solution: (a) We find directly J' the moment of inertia of F', either by the integrometer or graphical (e. g. by Čulmann's) method.

(b) On the axis u take two segments AB, AB' respectively proportional to J and J', and describe two semicircles on the diameters UB, UB' which intercept on the axis of w the segments AC, AC' respectively proportional to J and J';" and two semicircles upon the diameters WC, WC' which intercept upon u the two segments AD and AD' respectively proportional to J and

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(c) From O' draw a straight line x' parallel to a, and take upon a' and x the segments O'X', OX respectively equal or proportional to AD and AD, so that X is with respect to O on the same side as X' with respect to O'. Draw the straight lines OO' and XX' meeting at the point S.

(d) Finally, transform the figure F into the homothetical figure F, taking S as centre of similitude; that is to say, draw through S a series of lines cutting the contour of F in the points M', and the corresponding points M of the required contour Fare formed by constructing the intersections of the radii SM' with the straight lines OM parallel to O'M'.

This figure F is evidently the section required, that is to say, a figure which has given the centre of gravity, the orientation, the form, and also the moment of inertia, in the direction λ, with respect to x, equal to the given quantity J. In fact the ratio of similitude of the two figures F and F' is =4/J:'4/J'.

Note 1. When J' has been found, we can calculate directly the number

and then we should take O'X' upon a arbitrarily, and on OX we should take OX=μ.O'X'; we should then continue the procedure as above.

Note 2. If the position of x and the magnitude of J are not given absolutely, i. e. if the inverse problem is to be resolved several times supposing a and J successively variable, and if for the determination of J' (see (a)) we employ the graphical method, it is convenient to use the central ellipse of F'.

On this point, and for more details, see three notes that I have published in vol. ix. of the Rendiconti dell' Istituto Lombardo,' 1876, or my memoir, "Sul

We may restrict ourselves to consider the axes which pass through the centre of gravity O of F, on account of the well-known relation between the moments of inertia which have reference to these axes, and those which have reference to any parallel axes.

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