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368

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operation of elliptic turning is nothing more than that ofthe ellipsograph, or common trammel, with this difference,that in the operation of turning, the ellipsis is described bymoving the plane, and keeping the point steady, but in form-ing the curve by the ellipsograph, the plane of description iskept steady, while the point is in motion. The transverseA b e f is the same as the grooves in the trammel-cross, andthe line cdg the trammel-rod : here the cross and plane ofdescription move round together, hut fixed to each other, andthe trammel-rod c d g is held still or immoveably confined ;in the trammel, the board and cross are fixed together, andj held while the trammel-rod cdg moves with the points c! and d in the grooves.

j To set this machine, therefore, it is only to make C d equal; to the difference of the axes.

Figure 16.—No. 1, 2, 3, and 4, show the relation between| the foregoing diagrams and the chuck. Let k l m n be the| face of a board representing the plane, which is fixed to the; axis of the machine ; and let O p Q R be another board, madeto slide in the board kiks: each two points, o and k,

| l and p, m and Q, n and n, coinciding at this moment, elus: will therefore represent a wide groove in the board ; as thisi groove may be of any width, we may conceive the breadthto be very small, or nothing; it may therefore be representedby a groove, or by the line a b parallel to k n and l m, andin the middle of the distance between them. Instead of sup-posing the point d always moving backward and forward onI the line e f, we may suppose a circle, or the end of a largecylindric pin, moving in a very wide groove, tcvw, acrossthe slides opqe. Now, all the differences between these. diagrams and those in the former Plate, are only wide; grooves in place of lines passing longitudinally through the; middle : for the line ab is always conceived to move reci-i procally from one side to the other of the board klhkiand it is the same thing whether one straight line slide lon-gitudinally upon another fixed line, or whether a bar of anybreadth move in a groove of the same breadth, or whethera straight line in reciprocal motion always pass through twofixed points.

No. 1, shows the chuck, as in the first diagram of the lastPlate : No. 2 as No. 2, No. 3 as No. 3, and No. 4 as No. 4,of the said Plate. Any farther explanation is conceived tobe unnecessary.

Problem 2.— An ellipsis, a b d c, being given, to find thetransverse and conjugate axes.

Draw any two parallel lines, A b and C d, cutting theellipsis at the points a, b, c, d ; bisect A B at e, and C D at/;draw G ef h, cutting the curve at G and h ; bisect g h at I,which gives the centre ; from i, with any radius that willcut the curve, describe a circle, klnm, and join k l and mn;bisect k l, or m n, at o or p, and draw q o i p r, meeting thecurve in Q and R : then Q R is the greater axis ; draw s t atright angles therewith, meeting the curve in s and t, ands t will be the lesser axis.

Problem 3.— Any diameter, A b, being given, and an ordi-nate c d, to find the conjugate diameter of the ellipsis.

Draw C I perpendicular to A B ; bisect ab in f, and drawF II parallel to C D ; on F, with the distance fa, or f b,describe the semi-circle aib, cutting c I at i ; make Ae equalto ci, and draw e g parallel and equal to CD; through gand a, draw A h, cutting f h at h, then f h is the semi-con-jugate diameter.

This problem is useful in perspective, in the representationof the circle : for, having the representation of a diameter ofthe circle perpendicular to the intersecting line, and therepresentation of a diameter of the circle parallel to it,the former representative diameter of the circle will be a

, doublf

diameter of the ellipsis, and the latter will be ! i^this p ro ;nate of the same; find the conjugate diameter by^ c ur' eblem ; then, having the two conjugate diameters, ^ a y b®may be described as in Method XI; or the axi cU rf efound as in the next problem, and thencedescribed. _ A B and c p ’

Problem 4.—Any two conjugate diameters, a »being given, to find the axis. „ n dic u ^ r

Through d draw E f parallel to A B, and DI P er ^. a( jias 1 pto e f ; make d i equal to m a, or si b ; with the g( , t b/

describe from I the arc gT>l, and join I M, which ^ th®

a perpendicular, meeting the tangent E F at N . E ji ai' 1 'distance n i describe from n a semi-circle, e i f ; J° ^ j / f jf m, which produce to h and k; and join „ a t o an .,parallel to I m, draw l l and g G, cutting H e and K ^ wn*L ; make ir h equal to m G, and M k equal to M L >

G h and k l be the two axes required. c 0 l ,jug a f e

In like manner, if g h had been a diameter, its ^ joijjwould have been thus found : produce u G to E; „e i; draw e v at right anglescomplete the rest as before.

: produce H G to E ’ 63 to eij then draw 9

. !

This problem may be readily applied in perspecb^ f oU n<hby the last problem, two conjugate diameters wi W

and having the two diameters, the axes may be e llip tlC

this, and the curve be described geometrically by pas te "

compass, or by traversing the curve with an i 3board, or strong paper slip. _ . ^ gi^’

Problem 5.—An ellipsis and its foci f ana gto draw a tangent through a given point, H, in jjjgect

tb«

C p>., the

Join f h and g h, and produce the latter to ijangle i h by the straight line L H, and x, H is a tangecurve. o din&

This problem is very useful in masonry, f°r a c0 rv®’joints of elliptic arches. Thus, find a tangent m c0 nta ct ’at the lower end of the joint, and from the poift ° e pne ^draw a line perpendicular to the tangent; and thdrawn will be the joint. . - n. <F l< '

Problem 6.— Two conjugate diameters, A B atthe centre, u, being given, to draw two tangents tofrom a given point, e, without the curve. 0 j e<i o!l

First, let the point e be in d c, produced ; nj a l j, cato h c, and join ie ; through c draw C K, paralleting h A in K ; make h l equal to H K, and through >

parallel to A b ; find the extreme points F and G, byand draw e f and e g, which are the tangents req A p 0 r c fBut if the point E be in neither of the diameters, t j, r oU<?when produced, draw a line from the given P° in J t he P°the centre, so as to be terminated by the curv ^’j aC ohj u ® 9tion thus intersected will be a diameter = then fin

to this diameter, as in Problem 4

ire-'

Ctif ^ith

This problem will be very useful in the P e ^ s ^ s id& s

-the P

ire-

o» e ’.

the Utmost ex-i C, f ne ’ *° r araww S tne coin™.ceding problem? tDeSS J ^ diamefers bein S found b/

A b ci descmbe an ellipsis similar to a a ,ai

&***—*» „

in Problcm A ^ D and B c , are not given,

centre; throu-h n the ieP ° !ntE ’ W ' iere the 7 int <f^eet

curve in » • c * t le & lven point P, draw F E, t° citf'

tm<r A E at G J °r a *' andrB J parallel to v A draw E ®, f! ;then win x. f r, IK P araIiel to fb draw p h, cutting ® JesS ersemlalil. ^ greater semi-axis, and E H * e 165 ^

Problem 8 - * n, e f .

Through the angular points , ^

"• •- which sn

given rectangle, to circumscribe an ellipsis,its axes in the same ratio as the sides of the re

,le.