✓ (d2 — x2) = HI; and since CD2 — (DI

CI21H2; we have CD2.

the required equation. Drop the perpendiculars BL and CM;

=√(d2x2); consequently y = BL ± DH: hence, take

BL; and make Qɛ and QE' each DH; then E and E' are in the required locus.

If x be affirmative, and √⁄[ca — (a2o — b)2] = 0, x = — (6

+ c), its greatest value, and y = e = AB.

If x be negative, the above expression becomes

If y be a maximum or minimum, we will find

It may be observed that unless b = AG be assumed AC F

About 14 years ago, a young gentleman connected with Messrs. Bolton and Watt, and once a Pupil of mine, communi cated to me the substance of the above problem, and desired to know whether the curve FE (fig. 2 and 3) differed much from a circle: He said that they employed an iron bar ER, which played round a bolt R, to retain the point E in the curve. On receiving the solution which shewed that the curve was not a circle, he observed that the portion of it which was necessary for their purpose was sufficiently small to permit the bar to move without sensible strain on the bolt and gear; and added that the beam AC being 8 feet, the above system permitted a vertical motion of 5 feet, which answered every purpose. Messrs. Bolton and Watt have since improved the contrivance without, however, producing more than a comparatively rectilinear motion vertically.

VI. Let bf cut a given circle; and c be any point therein; and in it another point d be taken, such, that bc x df = cd2 : Then if any point p be taken in the circumference, and pc and pd be drawn cutting the circle in n and m, pd × dm × bd is = pc x cn x cf.

cn: pc: cb,

consequently pd: dc x cn :: pc: bc x dm:

or, pd x dm nc x pc :: dc: bc :: df: dc :: ef : bd: Hence, we have pd × dm x bd = nc × cp × cf. Q. E. D.

Cor. 1. Hence pd x dm × bc = nc x cp x dc.

2.

pd x dm x cd = nc x cp x df.

Several curious properties of the lines concerned may be deduced, and the hope of solving, geometrically, the ancient prob

lem, the duplication of the cube,' might seem to revive, by considering the point c in the centre of the circle as in fig. 4: in this case we have (bc)3 = pd x dm x bd; i. e. if p'dm' be perpendicular to bf and p'cn' be drawn, (bc)3 = (p'd)2 × bd = a given parallelopiped; and the side is to be found of a cube which is to this solid: But the difficulty lies in determining p'd and bd such, that a circle will pass through b, p' and fso as to make bc cd: cd: df. This theorem occured to me above 20 years ago and as it has gotten abroad, I think it proper to publish it myself in your valuable work.

ARTICLE VI.

PROBLEM.

By Mr. PETER NICHOLSON, Architect, No. 10, Oxford
Street, London.

To determine the nature of a surface described by a straight line moving along two other straight lines which are not in the same plane, so that the describing line may be parallel to a plane which is perpendicular to one of the given straight lines, and to find the sections of the surface, as cut by a plane given in position, and also the orthographical projection of the sections.

Let AD and BE be two straight lines not in the same plane. Draw the plane DEC perpendicular to one of the lines AD; and the plane BEC parallel to AD: draw AB parallel to the plane DEC at such distance from it as to be perpendicular to the plane BEC; through the straight lines AD and AB draw the plane ABCD which will therefore be a rectangle, perpendicular to both the planes DEC and BEC: let the describing line be KG which will therefore be a line on the surface parallel to the plane DEC, and draw the plane KGF parallel to the plane DEC cutting AD in K, BC in F, and BE in G. In the plane KGF draw HI parallel to FG cutting KG in I and KF at H: then HI will be perpendicular to the plane ABCD.

F

K

Let AD or BC= a, KF or AB = b, CEC, BDd, KH = , BF or QHy, and HI= 2.

Then, by sim. As, BCE and BFG, BC CE :: BF FG;

that is,

a : C ::y:

су a

Again, by sim. As, KFG and KHI, KF FG KH : HI;

that is,

су
b: ::

: 2.

a

Whence the equation of the surface is z =

cyx ab⋅

cyx
ab

Now suppose z abz cyx be an equation to the hyperbola; therefore the section of the surface cut by a plane parallel to the plane ABCD is an hyperbola.

a constant quantity, then will z= or

But to find the projection of the section of the surface cut by any plane upon the plane ABCD, the cutting plane being given in position to the given plane or to the plane ABCD; let BD be the intersection of the cutting plane in the plane ABCD, and let BHD be the projection of the section BID upon the plane ABCD draw HL perpendicular to BD, cutting BD at L, and join LI, then will the triangles BDA and PHL be similar, for the angles HLP and BAD, are right angles, and because PH is parallel to AB the alternate angles HPL and DBA are equal. Now HL represents the base, HI the perpendicular, and LI the hypothenuse of a right-angled triangle of which LI is an ordinate to the curve in the cutting plane, which will meet the surface at 1, and HLI will be the measure of the inclination of the cutting plane to the plane ABCD. Then, by sim. As, DAB and DKP, DA: AB ;: DK: KP, ba-by

aba —ŷ : KP =

that is

a

baby

a

a

-

Again, by sim. As, BDA and PHL, BD: DA :: PH; HL,

b.

ax + by-ab

d

α

Now let t be the tangent of the angle of inclination HLI to

which is the equation of the cutting plane; from which take the value of z, and put it equal to the value of z in the equation of

equation to the hyperbola, and from which is obtained y abt

b x ab't-cdrx

If x be supposed equal to b, then ya2bt

O

ab3t-cdrb = 0,

therefore make AB equal to b, the curve will pass through B.

If x be greater then b, the ordinate y will become negative, and will therefore lie on the other side of AR

till cdra become

b

to ab't, then will ya bt x

x = infinity; therefore make cdrx =

and draw GL through K parallel to AD or

BC, then GL will be one of the asymptotes of the curve. Ifx =

0, then y =3

a b2t abat

= a; therefore draw AD perpendicular to

AB and make AD a and the curve will pass through D.

Let a be taken negative or on the other side of A, and the signs of the terms in which a are concerned will be changed, then will b+ x

y = a2bt × ab3t + cdrx"

Lastly, if x be supposed infinite, the terms ab't and a'b't will

a bt cdr

vanish when compared with cdrx and abtx, whence y = ; abt cdr

therefore make AE

and draw EG parallel to AB, GE

will be the other asymptote: Therefore in both these positions the curve formed by the surface required and a plane will be an hyperbola, and also its projection upon the plane ABCD.

Let us now suppose the cutting plane to be perpendicular to the plane ABCD.