ELL

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ELL

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The method for describing an ellipsis, having two conju-gate diameters given, may be found in the Marquis del’Hospital’s Treatise of Conic Sections, translated by Stone.But the author of the Architectural Dictionary has chosento give the description and demonstration from a diameterand double ordinate, instead of two conjugate diameters, asbeing more readily applied in perspective. It is strange, thatthis useful method has been neglected by all English writersthat have fallen in our way. This property was dis-covered by the author, and demonstrated by him, many yearsbefore he met with the above work, in endeavouring to findout methods for describing the perspective representation bycontinued motion.

Method VI.— Figure 7. No. 1. Let A b be the greaterI axis, bisected in c, by the lesser semi-axis C d ; take two| rulers, c e and e p, of equal length, equal to the sum of thej semi-axes c d and c b, moveable upon each other at e, andI the end c of the rule C e, moveable upon the centre of theellipsis. Make the part p g of the ruler, e e, equal to thesemi-axis CD; now suppose CE and e p to coincide with eachother, and with the axis c d ; then move the point f from c,in the direction c b, until the describent G arrive at b : thepoint G will then have traced the quadrant db of the ellipsis.The other quadrants will be described in the same manner,by reversing and inverting the rulers.

Figure 8.—Another variation of this : Let A b be thegreater, and c d the lesser semi-axis, as before; take thestraight line h i, equal to the greater semi-axis, AcorcB;from i h cut off i k, equal to the lesser semi-axis, c d, anddivide h k into two equal parts at i; then place the jointrule c e p in the following manner, viz., make c e and e peach equal to n l, or l k ; the part c e being moveable roundthe centre, c, of the ellipsis, and the two rules c e and e gbeing moveable round e ; now let c e and e g coincide withC d, and the point f to coincide with c, and consequently gwith d ; then move the point f towards A, keeping it in thesemi-axis c A ; and when c e and e f come in the samestraight line, the point g will have described the quadrantof an ellipsis; the lesser axis (see Figures 7 and 8) is equal2 x f G, and the greater = 2 X CE + 2 X eg.

Demonstration. — Figure 7, No. 2. Draw G i parallel toC f, cutting c e at I, and draw e l perpendicular to C f, meet-ing c b at l ; produce c e to h, and make c n = c b ; joinH G, and produce it to It.

Now, by the general demonstration, accompanying thearticle Cylinder, we have cb ! : CD 2 : : A it X kb : kg 2 ;but by the property of the circle it h 2 = a it X kb; there-fore C b 2 : cd 2 : : it h 2 : kg 2 ; consequently c b : c d : :K II : k G, a property of the ellipsis.

But c E is the half sum of the two axes, and c I, or f g,equal to the lesser axis ; then I e, or e g, is equal to thedifference between the half sum of the two semi-axes, andthe lesser semi-axis; therefore ie = half the differencebetween the two semi-axes, and ih = the whole difference ;consequently ie = Eli.

Then because ig is parallel to of, we have ec ; ef ::e i : eg ; but ec = E f ; therefore ei = eg; and becausee i = e h, e g is also equal to eii : then since the angle e p gis a right angle, the angles PEG and egp, are together equalto a right angle ; but the angles egh and gep are alternate;therefore egh = gep; add to each of these equal angles,the angle egp, then will egh + egp = gep + egpequal to a right angle ; consequently IG is parallel to c k,and the triangles c k h and l g ii are similar.

Therefore Ch:ci::kh:kg; but CB = CH, and c d= c i ; c b : cd::kh :kg, the above property of theellipsis.

Method VII .—Figure 9. Find the foci E and > ^ aD dMethods I. and II. let the ends e and f of two rules ^ ft0 dek be moveable, the one round e, and the other rounlet each be equal in length to a b, the greater aXlS ’ ne ctedsecting each other at e ; let the ends i and Kby a bar, i k, equal in length to e f, so as to be ® carrI edround the points i and k ; then if the point i or K, be^ ;[ieround G, the whole instrument will be in motion,

point A will describe the curve of an ellipsis.

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it®

Demonstration. —JoinEi; then, because the ^ r ' an ®| ie Vft°and i f e have the two sides, I k and K E, equal /the ang IeSsides e f and f i, and the base i a common to both, s areike and I f E are equal; therefore the sides IB an er tyequal; therefore if = ab = eh + hf, which is a Pof the ellipsis. der the

Method VIII.—Another method will be found oi

article Cone , by the equal divisions and intersestraight lines. r poi

Method IX .—Figure 10. To find any number Jin the curve .— On the transverse axis, A b, descri ^ tb ecircle: take as many points in the cireumferen g jjjpti®semi-circle, as may be necessary for constructing 1 cU ttingcurve ; draw straight lines perpendicularly to ‘if the cenit, and let one of these lines, c d, pass through t ^ oD®let e f be any other perpendicular, cutting the axis jjjat f ^let C G be the lesser semi-axis ; find the point B, s 0 ;jit 3may be a fourth proportional to C d, c g, EE, and the ga inewill be in the curve of the ellipsis required . ia perp® 0 ,

manner, a point may be found in each of the otn m et .

diculars. The finding of points in the curve by 1 re adiW

being entirely in proportion, the whole may be v jess® 1

obtained, by making ik = CD, It, equal to g ^ I ftsemi-axis of the ellipsis ; join n; on ik make •equal to the perpendiculars; draw lo, mp, 11 J’ aJ . e tb®to kl, cutting i l, at o, p, q ; then io, i p> V’ u pon d 1 ®ordinates of the ellipsis, to be applied respective y 6 f\\Jperpendiculars, from the greater axis. This may or tb®be described by means of the proportional comp

CP

sector. „ x is,

Method X.— Figure 11. Let A b be the g reat ^ cribe tb®the lesser semi-axis. On the diameter A B, ® t j, e sei®'semi-circle A E b, and with the radius C D descri ^ tb^circle fdg; take any number of points, «> f> t j ie gem 1circumference aeb, and draw h c, i C, &c. c . ut * lD ® dr aW h n ^circle fdg, ati and l ; from the points h, h „ 0 jnt3. ’ ’h m, i n, &c. perpendicularly to A b ; also from the P f Jf&c. draw lines km, In, &c. parallel to a b , tb ^ f 0 r bm, n, &c. are in the elliptic curve. When the P ' t j ie otbthe curve are found, the corresponding points Den di®° 8 0half will be readily obtained, by producing the P n the ®to the other side of c b, and making the ordina eside equal to those on the other. . fp dia^-Js

Method XI.— Figure 12. Any two ^

Through P pQ ;

fp>tb®

A b and c d, being given, to describe an ellipms ^ p

found in any diameter, taken at pleasure. A ,. u ] a r t0p Q parallel to A b ; from d draw d f, perpen i d j s taiic®make D f equal to e a, or e b, upon F ; with t dr aW „describe the circle n d k ; through the • the tang®” j a tlines pen, t e m, ski, indefinitely cutting t ■ rC le» p ,’at p, t, s, &c. ; join pf, tv, s F,&c.cutting the' and®. ^the points n, m, l, &c.; also join e f, if ne , c . raet ers 11n n, m m, l Lj &c.parallel to it, cutting the the c

l l, &c. at n, m, l, &c. and these points wifi , uC ed to ^

of the ellipsis required: if the diameters are P ^ E 5l ’ g3 -opposite sides, at n, m, l, &c. and the distance_ jje oo ^

&c. are made respectively equal to t ^ eir .^ a points ’ponding distances e n, e m, e l, &c. ; then t r