IMAGE EVALUATION TEST TARGET (MT-3) fe /. &J [A 1.0 I.I 1^ ■tt 112.2 US m u 1 40 IL25 ill u 1.6 V] <^ ^ /2 / u/^. Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 f\ iV % V o^ W. I '4^ I f/u CIHM Microfiche Series (l\/lonograplis) ICIVIH Collection de microfiches (monographies) Canadian Institute for Historical Microreproductions / institut Canadian de microreproductions historiques vV nn Technical anrt Bibliographic Notes / Notes techniques et bibliographiques The Institute has attempted to obtain the best original copy available for film'ng. Features of this copy which may be bibliographically unique, which may alter any of the images in the reproduction, or which may significantly change the usual method of filming, are checked belcw. 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Les cartes, planches, tableaux, etc., peuvent dtre filmds d des taux de reduction diffdrents. Lorsque le document est trop grand pour dtre reproduit en un seul clich6, il est filmd d partir de I'angle supdrieur gauche, de gauche d droite, et de haut en bas, en prenant le nombre d'images n^cessaire. Les diagrammes suivants iliustrent la mdthode. 32 X 1 2 3 1 2 3 4 5 6 DOMINION SERIES OF DRAWING BOOKS. GEOMETRICAL • DRAWING III; Tin; I si; dl' SCHOOLS AND COLLEGES \;\ — C. H. McLEOD, Ma.E., Professor in the Faculty of Apiilitii Scifiu't-. l^cGill Umt'ersify. Montreal. N, / \ I \ sr - I V y / \ ,/ / \ / I \/ PUBLISHED BY FOSTER BROMZiN 5c CO., MONTREAL. 0.>.i I I. I „■■.■ M,jM.-:| . N3 preface:. The methods cmnloved nr.. fnr m.., proof, and althouKl. he con XuctLf ^ Kreater part capable of riKi.l student should in all els "nd"^^^^^^ " '"'"" '" '''" "'''^' "'" the drawinK. ""^erstand the method as well as execute and Macii^ie d;:s '^:^' .:^;:r, ::^:^":f '" ^" f^"--*"- ".'awinj stnution lines. Tlu'se slm ,. , '^ ''''''"■'"*^" represent con- but continuous pen n.u it ;;;.,7""''' ''' "" '^'""^"" "'' "«"^ br what i« re„uired The .tude ' '"^'^.^'^'-r"**''^"" """t '•'^«'ven heavier than tl e fanner and 1, T\^ '"'"'' "'" ''^"«'' '^""'""l.at lines. ItisanexcS ^eti t ";:;k i nT"-'^ 'T -"■'^'•"'•^-■' making the latter thicker t a. the onn '"^^'^'^"'""'''"''"''''''^ ""'••^' done with Indian ink and ^ rawing TeT [ll' ,""•'"' f ""''' '" made on light cartridge paper and sZld. ''-^"'"^'s -''o-M be possible neatness and accuracy! '"^"^ "'"' *''" '^'•'''"•'■^t The instruments required are the following ■- A small drawing board and pins ; compasses with pencil point; a Ti.e..,Hp„in.:,»;;r,;u",,tt'-,":r'"''' '* -"■"• -"n-- ..vJ;;;"';Tito';',r**r,rr It!; ■■■S;:; ■'■ "■"^' ■■«« b.'en made in the work a sinnll T '^\ "^ '"*'""" P^'Offess has to advantage, ^u\:,:uid^^tt 'e m^o;::: ^r^ T?;""^^'^^ "^'^" .-•ever, genera.,, be the mo. .LLL:1 f:^ of ^LS^-^dgT'li Prof. c. B. slJith, ^:2 n^'un^l^rih^ tr ILl^'L^SuL^ <" the selection of the problen>s and revision of the copy. " """'"''^ McGiLL CoLLEOE, Montreal, August 27th, 1890. c. H. Mel p:od. DEFINITION:-*. A vprHcal line is perpendicular to the liorizontal and would be represented by a line in the position c o, A horizontal lino is a level line parallel to the horizon an.l would be represented by a line in the position A B. Which forms right angles with anothe^br^^r;::::;::. ';;*;. j:;^ C D ,^ perpend.cular to A B and K I> is. perpendicular witl/refe.^nc e Entered acEording to AcT^Parlisment. in tiri^^,^;j^^ the II?r'.'*'I ""''!"'■' '""*'"'' '" "'« '"'"'« P'«»« <*"'l are everywhere the same distance from one another. try wncre of straight m!"""'" "*''"■*' '-P'-«««-e bounded by any nun.ber four^'sid^'™ S"-? 7 """^"^ '^PP'"^'^ *° ««»"•«« having more than roui sKlts. Pnesided figures are called pentairons • six sldp,! hex««on«; seven-sided, heptagon.; eight-sfded. oct««o„ " '„ ! Md d, non««onB; ten.sided, cleca«on« , eleven-sided, undecaKor. and twelve-sided figures, duodecaKons. "lecaRons , angle*!*"'" *'""^""' '" "'^""^ '"^"'^ ^1'"^' ^'«'- ""^ <-'qual NOTE.-A Inrge numberof the usnal .lefinitions h,r I without doubt well known to all who ..re likely >o use thi Office of the Min^iteT^f Agriculture. '6t..;v.;by"^:irMa:..^ ive been omitted OB they are is book. eraser. For irc required. The Sloyd to the com- luite sharp. ' thick, not roKress 1ms ifty be used quure will, ;ht edge to term in the lirecfion of assistance cl EOD. crywliere iiuiuber ore than »ix-side(I, 8; nine- uaffons ; id equal 5 they nre PllOHLEM l.-To BisKcr A Given SrnAKa.T Link, AH. B '/i From A, a? centre, describe an arc liaving a radius greater than onc-lialf A B. From B, as centio, describe, with the same radius, another arc cutth.g the former in n and b. Join, by a strairtt line, the roints of intersection a, b. The line Bb will bisect A B and will itself also be bisected at the same point. i'UOItliK.M ;{. To TmsKcr nii: Civkn Hk.iit Anci.i: A »(• With B as centre, describe any aro cutting the jjdes A B, BC, in a and b. With the same radius a.„l the centres a, b, describe arcs cuttinn the arc a b in c and d. The straight li„e. BI. BM, joining » to c and U, will tri.eet the right angle. ABf. PUOBLKM a.-To BisKcT a Uivex .Vxcii.E, AIU'. PKOBI.KM I.-From AGiVKN Poi.nt U. in a Stuaioht Link AC. TO Draw a P/:Ri'i;Ni)icii.AK to tiii: Lini;. With B as centre, describe any arc cutting the linoi A B, B€, in a and b. •rrom a and b as ocnlres, with any length as radius, describe ares meeting in I-. The straight line I. B will bisect the given angle ABC. Ta B Make B« eoual to B U and frnm the ,..)int^ « and b a>- centres, de.-cribe equal arc. meeting in o. The straight line BL, joining B to c, will be the required perpendicular PH<»IUiKM 4 (u).-Fr.)m A Given Point n IN A Sthakuit r.vp PRoni ..^» » ,. A C AN,, NKAU ON. ENM. CH- IT, TO I,HAW A pj, t v .,, , ^ ^ '^'''^'^Tb " H "' ^ '' n"'"^' ''°"^'' ^ ^^'^""^'^ ^ «'^'="' I-'^"^ T(i Tin.: flivi. V r.iv-r.- ■■^"" ' "A" A B, to Draw a PKHl-KNimin *» 1,. .,i.„ T i,- a B From the point B, describe any arc nbc and from » describe with a radius aB an arc cutting the formerarc in b, and from I. with the same radiu., again cut the iirst tiro in c From band «■ wi,h any radius, describe (wo arcs meeting in L. Ihe strii^ht line B L will be the required perpendicular. PROBLEM 4 (1,).-From a G.vkn Point R at tiik En„ ok a Line A U, TO Erei T A Perpendicular to the Line. <:< Take any convenient point c away fr"m the line, and with radius c B, describe the arc n H b. Join n e and produce it to cut the arc in I.. llie line Ii II 'fill bo the required perpendicular. ^ ..w... „ v..,,!^, juiivi \j vv ithout a ur A B, to Draw a Peri'eni.icui.ar to the Line. PI •.a. k. •It'C From € as centre, describe an arc cutting the line in the points « and b, Jor a <■, describe the arc € c. "■" •■^reHnlL'e^SZ c?e.^"'' "'""^ "* ^' ^-""'^ " --'' ^ -*"- ""> .loin Cc. cutting A B i„ I,. v I. will be the required perpendicular. Drill Join PK From Theli GivKN Line ach other ia c r. (JiVEN Line I'ENUICULAB rc cutting the Ml;i.nN(iINA(MVKNAl,,,A(ENrL.NEAn. AN.. MAKIN.i K'e''ny unit of measurement, !l^. .( »- A * oC D Draw a line <' D imrallel to the given line A B. Divide <" I» in the desired ratio us in the flr?t method. (0 D ibnuld have either a xroater or leas lenitlh than \ B. preferably greater). Join A ti> <' and B to D and produce these lines to meet in n. Join i»b, nc< n, without Puonut'iNo THE Latter to Intersection. a B /: II Draw a straight line through C, meeting A B in a and C B in b. In any convenient position draw another line paralled to ab, meeting the two given lines in c nnii d. Join ad and through Kdraw a line Ee parallel to C U. cutting ad in «. Through ndraw ef parallel to AB, cutting vd in f. The line Kfli will be the desired line. With B and C as centres and any radius, describe ares of circles meeting the sidei of the triangle in the points a, b, e and r. With Band E as centres, describe with a radius eiiual to Ba orl'f, the arcscd, biri make the chord cd equal to the chord a b. and the chord h|r equal to the chord e T. .Toin Dc and join Eir, and produce these lines to meet in 1.. DEL will be the required similar triangle. Thrt Abo Join AL PR PROBLEM 12.— Through a Given External Point C to Draw a Straioht Line to make a Given Anoi.e I) with a Given Line A B. ThroHgh € dr.iw n. Hnn «!«, p.im!!n! to \ H (Prnh. S.) From C draw a line C'li, making an angle withCe equal to D, and meeting A B in L. (Prob. 11.) The line € Ij will make the required angle D with the line A B. Proc Bisc Atl AB Construct c. itiog the sides arcscd, bg; ; equal to the DEIiWillbe TO Draw a H A Given I'UOnijKM I :l— To CoNSTUUrT an E A and at 11 make an angle equal toC Make D L eciuil to H and join l< \'.. The triangle n I^ E will be the roiiuircd trianiilo. IMtoniiKM to.— To Constrit(;t a Thianoi.k IIavinii Two Sn)E8 EyCAL HKSI'KCTIVKI.Y TO A AM> H, AND AN ANCil.K Kl^UAl, TO THK GlVKN ANIil.K C, SlTUATKD Ol'l'OSITK TO ONK OK IHK GiVKN SlUKS. meeting A B T~7 & Produce the base B A and oonatruct un angle c Ad equal to the given angle C (Prob. 11). Bisect the .angle d A 8 faj- the lino Ac (Prob. 2). At B make an angle K B b equal to e A f and produce A e, B b, to meet in I<. A B L will be the required isoscles triangle. B D ___ £ Draw the lino D E equal to A and at B make an angle E O L equal to the given angle C From E as an a centre describe an arc with a radius e(|ual to B. If this arc cuts the line B L. in two points H and It, the two triangles It M E, D L E, will fulfill the conditions of the problem. If the arc merely touches the line B L, there is but one solution, and if the radius is too short to reach the li'.ie, the case is an impossible one. ■^-T '•"'»•«•■•:>• 17. To I)„.w A Mkan runvnunas.u. n. Tw„(i,vrv I'IMS .\ AMI li. A .£ i»ii - I'-. JL .'''a .."(k «."' «< In .be li,,,. ,. br „„.1<. „ „ „„„ ..^ ,^,„., ,„ ^^ ,,, „ rc»„eo,ivelv ''i«o„-„,|,„,„eterdi.scril,o„scM,iicircle. ii- '■*«!>. ' • "r;;!,;;i[^';;;;;;:;:;:[-^;»:; -r -« ""> ««■»!<"-•- >" ^■ xbi. ,,„„ wu. be .ho "' lXuTjl\i::^tV' '" •^•""" "' "™'' '° '■'« reoh.n,lel„.vi„, adjacent ,iJo, IM.o.U.KM ,s._T.. F,x,. . Thmu, Vnu.onru,s.u. .u Two (i.vKN l-lNi;s A AM) H, 9,f ' / / J. ^. -r -tf ffom nn.v Miikn bd ''nil, )l on Tliethinl ell NiHK.— If point bdniw two ?tr,ii({lit linea n b, be Cual to A „ncl bo e„u„l to B. M„ke also o|r equal t„ A or bd Fromnny point b draw two stniiichi lins^inb, htt, Makohifc.iimllo Aiindbft,|,ii| loll. MuU« «!=,; 1.^. . ,. « . Join M-and throuKh e draw ed .huhIIoI ^^^^^t^mlCT'l "^l k . be 11.0 required third proportion^. " '''" """" "• "" "'" At Mak Pin From any „ojn, „ jraw two stralKht lii,es Ad, HO .Make «b equal lo V, and nc equal to 11. i|„|^„ ,,.,,, i.rf „„„„, ,„ - Join be and draw do parallel to be, oiUtiLVi , !«.? . ■„ u required fourth proportionu- ''* """' *■ •"•" '^'" ''« "'« ejraction wii. .ive a nrh';:^;;;;:,;;;^ K^ :!:'^,;":;;„„3vTr"':; yfaorcr than tliu ({jyen lines. ' •"'"■■-fijr sitaste-l U, ee and Draw Join • On b< PIU)HM.;m ao.-T<. Divim; a STRAi.iMt I,i,m: ,\ M into E.XTHKMi: ANij Mkan Hatio, (i.e.) IN Mkdiai, Section. ••« I'UOIIIJOM 22.-T0 TlXl. run SylAUK Wool OK A (ilVKN I,IM: II, Hkkkkrki) to any Unit A. "■■K. At oneemi of the line A erect a perpenUicular A a and male. itn,.ml tootie-half A n. Join n li. Make •»»• e'lunl to » A an.l B I. equal to B b. The point I, vrlll Jl,|.le tl,o ||n« A » in oitremi ai.d mean ratio, that is, A L i E, B : i L B 1 A B. ■ ! h B a I t \ •iC L-.. PUonr.EM ai.-To Dictkhmink any RKgiriHKi, I'owii. of a ((ivkn Line B, thk Unit \ Bkinii Oivkn. Make i» b c<|iial to A and b r in n b priKiuccd chuuI i„ B. l>e.cribe a nrcle on »e a, a ,li eter an,! „t b erect a perpend eular, meoiing tho »e..ue,rolo .n H b I, will be the square n.»t of the length H l.,r (he u«' ,»! ^K--. ''k--.; ^•>^- IMloni.KM a:i. Upon a Given Link A n to Consthp, t a Sg,-A„,. fi : / B» icr B» B^ """ 'laehThe""' ""' *'""" "'"""^"'^ '° * ""•» "' '"d »t right angle, to Join «c and draw c I. at right angles to «c, meeting «b produced in L. bL will be equal B-' to the unit A. «« m ».. » n On bcproduced make bd «oual .„ h ... „..^ .j,„„ .« .._.,,, , . , ^ ~, '"""r" " ''"''''"'''°"'''' *'*'''"'• ^ '''''"'''="' ""^ *''*''•"''''* «• bj. produced in M ^^^ni ^.^'i,; ,;-;^:j^£':^^^;^^;;;i I •^'■"■'"^X^""''"""''''"*'''"''^"'''""'"'"'-*'^^ b O, etc.. may be obtained respectively equal B^, B\ etc., etc i 1 « n . . ' • •"'■ I Join-HB. A 1MB will be the required .quare. At A in B A erect a perpendicular A » (Proh. 4 4) and cut off A I, equi.l to A B. JU A TO c'" '" •'"'^■^™*'""'' ^^'■•^■'^^ '^ ^^^•" "AND A niA.OXAI, KguA'r. C Bisect A B (Pr„b J) in », ,uul M « erect the perpen.lieni.r b« o. Make a I. and n M e.|u,il t(i n * .Hnd n n. ■Join Al, I, B. R M. in A. f. f„nn (lie re „.ired «„.„re A I, K !W. I'Kom.KM 2r..-(;,.o.v A (;,vi.;x IU«E AH t„ Const.,,-, t a Tm- T,,e„ -,^'^^-;^e;;f^»rj^.h;y..n^ centres ;ith ridii ec,ua. respective.. Join M y, I. y, complet ing the req uired pnrallelogram l.O M N. TO Another Thafkzr-m AMC'Ik J( U At AniakollieanglenAeequnl toC'(Prob. 11.) Bisect AB in » ^°" '^ngTb!"""'^ ^ """' I'erpendicularM,, A B and A e respectively, meet- From b as" centre, with radius b A, dc,«cribea circle li ,* BM ''"""'''^t^,}^glCV'^::;^^ A" a-'»tthe gi;e; distance D fro„> it, """ '"alfitu.* ?»'■ "'"' * " " ""' '"'^■^' ^•-'"="' ""«'- -I"'" to C and the required .Make a^ angle^ l!'^",^:'"^' '« »i » (Pi-'b. 11) and having ,l.e containing .Ide., -■! Si. .. .Ts. r.,n„i ,,, K A, A •>. respectively. ' ^rlT^^.l^^'^i^^ "'"' "^'" ^'>"»> «^'i'ec.ively toB«' and »«•. describe Join the point O to M and Sf. MW O N will be the , equired trapezium. Pro Joii 1 IIAVIX(; SiDKS' :a(!o.val i;(iUAL PROBLEM 2«.-lT,.ox A r.ivKx STRAKiHT Line PO to Coxstbuct A RlXTlI.INEAR FKiURE SIMILAR TO A GlVEX RECTILINEAR Figure A B C D K. N il respectively to lunl respectively ME nc B will bo ciual in nreat(. tlie originnl scvcn-Mdea6gHrc,a;id a Iciur- sided figure „( e(,ual area may bo oblainod by joining l> M, etc., and from that again a triangle of equal area obtained, but a better conditioned triangle will be formed by recommencing the process of reduction at B, joining B i> and replacing the lines B<\ CD, DK. by construction similar to that above de.cribed, by the line EL. The triangle ME I, will thus be the rc.uircd triangle. PUOBLKM ;iO.-To Coxsthi. t a Tuiax.m.k win. h shall hi: KyiAL IX AliKA TO AXY TlilAX.a.E OKK AXI) SlMU.Alt lO ANY OSIIKK Tkiancli: a BC. li AND LQl-AL Join A C and A D in the given figure. Upon FO construct a triangie POL similar to ABC (Prob. 11). On FL con- struct a triangle PI. M .imilar to AC D; and on FM construct a triangle FM N similar to A I> E. The whole figure FO L M N will be the required rectilinear figure. KO PUOBLKM au.-To CoxsTKUcT a Triax(;le Kqlal in Area to ANY Given Polygon ABCDKFG. containing sides 1 IK', describe um, M /, A B Product) any side A B. Join A F and througi. « draw « b parallel to F A. Jni„ F h. The six-sided figure bFED<-B will be equal in area tothegivensever.-sHcd figure. Now join b K and through F draw F M parallel to E b. Join E M. The five-sided figure Find the point « in 1»F so that the altitude of the triangle OnB shall be the same as that of ABC Join » E and through F draw a line parallel to a E to meet l» E produced in b. Join nb. The triangle I>nb will have an altitudecqual to ABt'and an area equal to D E F. Now find a mci.i ri-oportional EM between A H and Bb (Prob. 17), and on EM construct (Prob 11) a triangle similar to ABC. E M N will bo ihc rec.uired triangle I'llOIUii:.-*! .•{|.-ToDivii)i.; A GrvEN Tkiaxgi.e ALC i.vio any NlMllKll OlKlJUAI, ')■. Fnoi>OHTlOXAI. J'AUTS IIV L1.VE.S DkAW.V I'ahali Kr, TO One c' rirE Sides. ' .-;/ Divide one of the sides of the triangle, as I.<' into the iironortional parts (I'rob. 9) ii. a.'.b and o. Describe a semicircle on I- V as a diameter. At the points a, b and c, ereet the perpendiculars from the side I- «' to meet the semicircle in d, e and f. With I. as centre and I.d, I, e, I. fas radii, describe the arcs euttinw the side A<' in in, O and <{• From the points M, O and Q., draw lines parallel to the base A<^ thus dividing the triangle into the required parts. PUOIJIiKM ;ia. To DiviDi: a (iiVE.v TKiANtii.i; A IJ C tnto axv NlMUKH OK ECHAI. OH I'liOPOHTIONAl. I'AKT.S HY StH AKJIIT Ll.NES Uhaw.v iHoM A CivK.v Point I> in One of the Sii)i>. Diyide'the line A B in the required ratio in sr, f and c. Join IX' and draw the lines is Ij. f M. eST piirallelto !»«'. The straight lines l» I-, » M, I» N will divide the triangle in the given ratio Air: irf>. fe: eB. PROIJLKM ;{;i.-To Diviue a Ciivi.N I'aham,ei,oi;ham A KC I) into Tno I'AiiTs nAviN(i A Given Uatio to Ka( h OTiiEit, iiv a I-IXE DHAWN l-liOM A (JIVKN I'OIXT K IX OXE OF THE SlDES. PH 1 Divide A B in c in the required ratio. Draw c d parellel to A l> or B «'. Bisect cd in e, and draw the straij,'lit line E I. passing through «>. EL will divide the parallelogram in the isivcn ratiu. IMSOIUiKM a4.- To Divide a Givex I'oEV(;f)x ABC lX(i AXY Xr.MBEU OF JlylAL OH PKOPOHTIOXAL Sthaicht Lixes Dkawx fkom One of the Axui.i 1> i: I' HAV- Pahts by ;s F. Join Bisei Bisei PR Draw the trinnile .1 F J (Proh. 2<>) eqnal in area to the given poly Divide the ba?e of the triangle in the required ratio in the points Join FB and !•'<'. Through K an i; F HAV- Pahts HY s V. '■> md b, eduction at Ki PROBLEM 85. -To Fixn Tin: CKNTiiM am. Hadics ok a Civkn ClR( i.K A BC l>. D Join by a straight line any two points a, b, in tlio circumference of the circle. Bisect a b and draw a line at right angles to it, through tl.o point of biwction, meet- ing the circumference in 1> and «'. Bisect «' » in I., which will be the centre of the oirole. PROBLKM ao.-To Descrihk a Ciik i.k tim; Cim l-.mikuence ok Whicji Shall Pass Thhovgh TiiHEi: (;ivi:.\ Points A, B and C. Join A to B and B to O. Bisect the lines A H, !l f. I,,- ,|,« ,,errcndicu! irs ah. «lc. meeting in li. A circle described from I- as a centre, with radim I. A, will have the point? B and f also contained in its circumference. PROBLEM ;I7.— To Ukaw a Taxukxt at a Given Point A lo a ("iRci'i.AH Akc C a B. ng. 1. Fig. 2. (Fim ihiUmU t'ig. I.)— Take points a and b e((ually distant from A. The required tangent will be a lino drawn through A parallel (o ab. (Sfcimd Mitli'iii, Fig. 2.)— Take any point a. Join A a and bisect the arc A a in b. \Vith A as centre and Ab as radius, describe the arc cO. Make the ore bd eriual to the arc cb. The straight line E. M, passing through A and •!, will be the required tangent. PROBLEM ;»8. To Draw a Taxoent to a Cihcli; BTI) krom a Given Point A Without It. Find the centre E.andldraw the straiaht line EA. Upon EA as a diameter describe a semicircle ELA. cutting the circle in I.. The line joining A to li will be the required tangent. ISOIUilOM .-»». -To DhAW A TaNOENT to A GlVKX ClU( lI.All AH( Dm lUOM A GlVKX EXTERNAI, PoiNT A. WITIIOIT fslXCi THK Ckmri; ok tin: Cihcii:. -■•-v.r "ii! \ Draw a struight line throu?!! A cultinr ''u' iirc in B i.Dil V. Pn.duco <: A, making A n e(|u:il to B A. On <'a dc; . oc :i semicircle. At A ilrawtlie perpenaicular Ac to the line <'», meeting tlie semi-oirciimforcnce inc. From A. with A c as r.iiliu.-', de-oribe an arc, cutting the given arc in 1.. The line A I. will be the rniuired tangent. PUOBLKM 40.— To DitAW iiik Common Ta-xuknis to Two Given CiRci.Ks ABC. DKK. riraw any radius O A of the larger ct.-ole and cut off the part A a equal to the radiu? of the smaller circle. With O as centre and O » as radius, describes a circle » b c. From the centre H, draw H b tiinuent to a be. (Prob. 38.) From the points H and *i, draw iierpendiculars to the line H b, meeting the circles in H and I<. Ihc straiKht line M 1. will be a tangent line, atd a similar tangent line may bedrawn on the opposite sides of the circles. To Di-eiiv a Tiingent Crusshuj the Line of Centres. From H as a centre, witn a radius Hd equal to the sum of the radii of the two circles, describe the nrocd. From O draw a tangent to the arc e E. At any yoint b in the line n E make an angle D b a equal to the given angle F. From the point c drop a perpendicular cd on b a, cutting the circumference in e. At s dr.Tw -v lir,e M I. i,erpendieu!iir tu ce. M li will be the required tangent. '^ sting the circles le may be drawn 'res. idii of the two indiculars H e, les in Q and P ler tangent line I of the circles. IRCLE ABC VK DR. angle F. irence in e. 1 I'UOItLKAI 42. To Dewhihi.; a CiHdj.; IIavinm a Oivk.v Radiih E and ToL'(inN(; Two Uivkn SiiiAKiiir Li.nks A B, CO. IMtOBIiKM 44.— To Di:s(HiiiK Two Ciiu i.-jm with Givkn Hadii V AM) I), TOIIHINO EA(H OtHKK A.Nil) A UlVKN LliNK A B. At liny points » and d erect perpeniliculars wband cd to the lines AHandt' O respectively, and both equal lo the given radius E. 1 hrounh b and c draw lines parallel to A l» and <' I», respectively. The circle dc-cribed from the point of intersection M as u centre, with radius E, will touch the given lines A H, <' l>. Draw any perpendicular »<-b lo the lino All, and make «bci|ual to l»aiidi»c ctiual to <'. ThrouRh v .Iraw cd parallel to A it, and fmm the centre b with radius bd cciual the sum of the Riven nidii <' and l», deferllio an nn cutting the line €'d in d. About b and d us centres, describe the rc.iuircd circles with radii respectively ei|ual to l> and V, PROBLKM 4«. To Desihibk a Cirii,k Touching Two Given Sthaioht Lin'es AB, A1), and Containi.mi a Point C i.v One of the Lines. I'BOBIiKM 15.— To Draw a Cihci.e, the Cihc i-mfehknce of MiiK I! sjiAi.i. Pass Tiihouuh a Civen Point I) and Tout ii a Given Strakjiit Line AE at a Givkn Point C. From A as centre with radius A<', describe an arc cutting A B in ii. At «' and « draw perpendiculars C «• md nb, to A » and A B respcotivelyi meeting in e. From the centre e, with the radius e. PKOBLKM 47.-T(. Dkaw a Circlk of Givkn Radius E and Touching Two GivKN CiHCLKS AB, CI), havino Centres a AND c. Respectively. With a as centre and a radius equal E plus the radius of A B, describe an arc With c as centre and a radius equal to E pluD the radius of C D, describe a second arc meeting the I'urmur in b. Join bn and be. The circle described with h as centre and E aa radius, will be that required. PROBIiKM 4H.-TO Descbibi; a CiHri.K of G WHICH Sham. Toihh Two Smai.m.h Cim i.i Include Them. iVEN Raduts E, ;s AB, C I), AND From the centre a, with a radius equal to the given radius E leas the radius of the circle A B. describe an arc c e. From the centre b, with a radius equal to the given radius E less the radius of the circle C », describe an arc ed, meeting the former arc in e. From e as centre with a radius equal to E, describe the required circle touohing the given circles in f and g. I 17 ' / i Join Join From •I 1 I PROHIiKM 4».— To DiisiHiHi; a Ciiui.i': Toichint, Two (iivK.v PKOIUiKM 50.— Kkom tiii; Tiiui:i: Civkn Points A, U. V, as Ciiicr.Ks Ht', l> F, AND Oni; oi' Tiiiim in a CJivcn Point A. Cionirks, 'k) I)i;s(uifii: TiiKiii; ("iiiii.i:s ToiKiiiNd Kaiii Othku. Fi^. I. 'Xl Fig. S. Join the given points and bisect the angles A. II and C by lines meeting in a. ''-•omadrop perpendiciiliirs »b, nc, Hd on the lines AB, Bt', V \, respectively. With A as centre and A b or A «l as radius, descrite a circle. With B as Centre and Bb or Be as radius, ilefcribe n circle which will touch the former in b. With <' as centre and <'cor<'€l as radius, degoribe a circle touching the two former circles in d and c. Join the point A to a the centre of the circle CB. In this line, or in the line produced, make Ac equal to the radius of the other given circle D P. Join b, the centre of the circle DE, to c, bisect cb in d, and draw the per- pendicular d Ii, meeting A a produced in I4. From Ij ag centre with the radius I< A, desoribe the required circle. PROBLE5I 51.— To Inschibk a Squahk in any Givkn Thianole ABC. N «- From the point B draw B a perpendicular to A t'. Draw B c parallel to A V and equal to B A, Join A c, cutting B € in Ij. Draw I< H parallel to AC, and LO, SIN parallel tuBa. IK Ij O N will be the required square. IMlomjOM 52. To l.NscKii.i.; a SyiAUK ix a (iivMN TuAPMzniM PHOKMCM .",. To DiisdniiK a Trianom-; Simii.ak to a Givkn A lU' l> WHosK AiUACKNT SiDKs A B, AD and also IM\ V l> ABK EyuA]. TO Onk Anothkh. B a M TitiA.\(iLi: DKI'' AHoiT A (iivK.N CiHci.l-; ABC. Join A V and H l». Draw B n iiariillcl to A V and maki; it C(iual to II It. iJraw A n cuttintc II I' in 1.. Through r draw a lino iiarallel to A 4', nieetinK A It in M. Draw I< « and M X parallel to 11 n and ioin Sf O. to complete the rociuired square n I. it 51. PROBLEM 5»,-To iNscRiDh; a Crm i.k in a (Iivkn Tkianci.e A BC. B A N Bisect the angles A and € by the lines A c, «' c, meet ing in c. Drop a perpen.licular c W on A V, and from c as centre with c W as radius, describe the required circle, touching the sides in I-, M and N. IMJOBLKM 54.-ToDiisi;Hii(K A CiH(L|.; about A GiVKN Thiangi.k. (Si;i,- Proulio.m ;j)i). Produce one xule „f the given triii.iKJe K F and from the centre n of the eivcn circle drawniclii making the angles <•» A, Anil, respectively e,|ual to the angles c i l> and n K b. At the points A, II and <', draw tanger.ti to the uirule, meeting in L, M and N, and forming the required triangle. PUOniiKM .->«. In a GrvKN Cihci.k to Lv.scrihe a Tkianolk SiMii.AK TO A Givkn THiANtiUo ABC. til V Draw a straight lino a b tangent to the circle at a point I.. At I, make an anple a I. c equal to II A <" and an angle b Ld equal to B V A. Produce I, c and I-d to out the circle in BT and M. Join N M, to complete the required triangle I, M N. '() A GlVKN PKOBM'.M 57- In a (;ivi;n Cim i.i; w itii Ui:miii: A, lo Inhcrihk I" A HiKiTl.AH IIl:XA(i()X. I ] Riven circle to tho iiriKlos t and W, and Tkiancjle UOIti.EM SO.-ls A GivuN Ciiui.K AC II n. Diis. rihr an 0<'TA(i(>N, ■jC Drawnny ilinmeter and from it^ extremities L, n, mark (ff chords l Bf, to, M P, equal til tlie radius of the circle. I. N <| M P O will be the required heiagon. 1 *VA. Draw any diameter C'«B and from the point i in the circumference of the circle set off the chord C b equal to the radius <'». Bisect V b by a perpendicular, cutting the circle in d. At d draw the tangent l.d ft, meeting the diameter BC produced in I,. From a as centre with a I, as radius, describe a circle and produce the diameter of the given circle t'B to meet the circumference of the larger one in O. From t and O as centres with ta as radius, mark the points M, N, <|, P. The lines joining these points will be tangent to the given circle and will form the required hexagon. Draw the diameters A B, «' D, at right angles to each otiier. in Ij, M, N and O. Join A L, Ij V, etc., to form the required octagon. Bisect the quadrant.'! PUOniiKM .->«,— To I)K3(RIBK A RwiLI.AR HkXAOON AbOUT A Given Circle ABC. PROBLEM eo.-UpoN A GivKN .Strakuit Link A » to Constrix t AN OcTAliON, At the extremities A and B erect perpendiculars Ac. Bd, to the given line. Produce A B to a and b, and bisect the anglea aAt-.bBd byAfandBe respeo tively. Make A K. and B (| equal to A B. Through !. an.i Q .iraw !. !S, H F. pamllcl Ui Ac and equal to .i B, and through M and P draw lines parallel to B^ and A I, respectively and meeting A c and B d in W and O respectively. M N and O P will be equal to A B and on joining W O the required octagon will be completed. I'ltUHLKM «1.— To Ixs< Hini; in a liiVKN Ciaii.i; anv RKiiin.Aii I'ol.YliON. Let it be required to inscribe a penlHgon Draw any diameter A B and divide it into ai many equal partJ a« the required figure has sides. From A and B as centres, with radius A H, dcBcribc arcs raeetinKiin i». Join n to c, the se'iond iioint from A on the divided diameter, and produce the line to meet the circumference in Ii. Measure chords equal to AI,, dividing thi circle into five equal parts, and join the points. PROBIiKM «2.— Upon a Givkn Strakuit Li.ne A B to Co.v- STRIJCT A Rlor.l'LAK P()l,Y(!ON. Let it be required to construct a regular heptagon. Produce A B, making B a eciuiil to A B, and describe a semicircle on A a. Divide the semicircle into seven eiiual parts (I'rob. 01) in f, P, e, etc. Join B to P the second point from a. Find lliu centre of tbc circ^Io passing through A B P and describe the circle. Measure arcs equal to AB or B P. from P. Tliose will divide the circle into seven equal parts, forming the u luired polygon. The points J>, M, Bf, O, should be on tlie lines drawn from B through b, «.', I «•».— Tn I)i\ii)i; A CiiK i.i: .\ l»«' inio any Nitmiii;r OK ('()N( KNTKIl' I{lN(iS, TlIK AllKAS Ol' WMK II SIIAI.I, IIAVK A (iiVKN Hki.ation to Kacii Otiikii. I'KOIIIjKM 05 iai,~ (Si;ro.M) Mi:i iioM.i— 'In CciNsiiirc iiAviMj riM': A\H> A H. ("I» (liviiN. an ICl.l.ll'^E *.— 4 Drnw tlie radius nA unci divide it intu parts having the iloaired ratioi in the points b, o, (I. Upon n A as diiinieter, describe a somicirote ami erect perpendiculars to the line An from the points b, p, (I»».-iTiiiiii) Mi.tiidd). To (.'oNvimct an I;i,lii'.-. Draw the diameters at right angles and bisecting eoeh other in the point K- From <' or I> as centre, with 8f A or u B as radius, describe arcs cutting the major axis in the i oints a and b. Those points will bo the foci of the ellipse. From n and b as centres, with radii Be, Bd, Br, etc., describe arcs above and below the major a.xis. From the same centres describe aiso arcs with radii A c, A tl, .1 c, etc , catting the former arcs. The arcs with radius Ac meeting those with radius Be etc The points in which the arcs meet will be in the circumference of the rei|uired ellip.'e. O Construct the rectangle A FCOB. Divide XK. EB, into any number of parts (for siiniilicity use eiiual parts) and divide A V, O B similarly. Join O, the opposite extremity of the minor axis, to the points of division n, b, v, etp,, in \W, r.r.d pr("!ii,-'e the liiier t'l meet lines drawn from <' to the points of division in A P, O B. The iiointsof intersection J, k, I, etc, will be in the required ellipse. I'lioiiiiKM en (c). -To CONMTRiriT AN Kl.MPSR BY ClHCL'I.AR Am X, MATEi,»- IMlOllIiKM The rioint m whore B to out.-, i B will be the centre of the remaining arc K A. t'raw the other (luailriiiils xiiniliirly. A closer approximation may be obtained b> ling tlr es A«», AK into four or more parts, and obtaining f U« centres ui the aroi i gim.mr manner. I'ltOllliKM «i» ((!».— To CoNsiiitJCT AN Ei ' ipsk Appkoximatki.y BY Mkanh of Circular Ah( s upon Givkn Axks A B, C I). C Mnkn E It and E b eqnal to the difi'erence between the axes A B and C It. Make also Ed and E c equal to three-fourths of this difi'erence. (Ea= AB-€ D and Ed = ^i En), itoin * and b to d imd c and produce the lines. VtiitD d and c with radii equal to d A or c B describe the arcs e A f, g B li and from b and H with radii eciual to b (', or M I> describe the arcs TC (;, h D «. These arcs will meet in f, K, ta and e and form the required approximate el'ipse. This method is described by Mr. B. F. LaRue in Enaineering iVeio*. vol. xixiv., No.l7. Draw any two straight lines a b. «■ d, acrojs iha ib|lt|ii'« «hrniorindund 0. Join bd and br, cutting the cuire In the joints e and t, which will lie (hs iKnRent points. Draw the tangents »>I., Ef.H. tK.NCE OF . PUOUIjKM <»».— To CoxsTiiur t a.\ K(;(i-.siiAi'i;u OvAi. (».\ a (iiVKN DiAMKTEK A n. l>i(OKI. lueotinK n4' i Malio Dcuii I l»d,on t'Band Us produolion,ec|ual l"<'b Tlioliii'rt d rawn throunhd, parallel to A II, istliedi! of tbeiwrabola. IMvlile Iho iixis to any parts in the points e, v, f, g, etc.' lines pill Del to AB. From c us contn- .viih de aa radius, describe arcs cutting the « in e' aim <•". {■iinllurly from v ;. centre, with do. df, ttg, etc., as rmlii, the piinilli i iinrs in €■ iiiid c". 1 and I > K' and H"< <'< 'lh« polniB so fijiiiiii will be in the curve of the i. irabohi, whiel hand throuun Ihvm. J III b. uid V is the focus through these draw 'illel line through ribe arcs cutting 'o dr»wn free. I'KOIIIjKM 71. The BA.sEoitDoim kOudinate A»; nktheAxis OH llEKiH CI) OK A Pahaboi.a FJi inu Oivdn i uka Willi: ClHVE. Bisect AB in n, Drnw ab at right angles to AB and equal to a.4 or aB. Juin Ab, Bb, and produce them. With A and B as centres and AB as radius, describe the arcs Be, .Id. With a as oontre and a.V as radius, describe tlie semicircle AMB. Wilh baa centre and radius br, de-Ci-ibe the arc dl,c, completing the required liguro AMBI<. 1 1 raw Aa and Bb parallel nd equal to <'D, and join ba. Divide .\.i' and <'H into ar. ~ number of equal part.5 in ti.c ;>oint3 c, d, etc. Divide also Aa and Bb in? > Ihesame number of equal parts in ni. n, oand J. k, 1. Draw lines through c. d, e •., parallel to *'D or perpendicular to AB. .Iniii n to the points J, k, I and m. v, o. meeting the vertical lines in P( q. r. a, t, ii» which will be poi \ta in the required parabola. PKOBLKM 72.— To Dhaw an Hypekhola UAViNfi Given xni: Douiim: Ohdinatk A B, the Hekjht or Abscissa DC, and THE Vertex K. Constiuot the reotnnRuliir figure ABbn. Divide A It into any number of equal parta in c, d, oto. Divide A a and B b into the same number of equal part? in m, ii, o and J, h, I. Join the points in A B to K and those in A n and B b to <'. The pointB of intersection ol these lines will be in the required hyperbola PHOBMOM 7:J.-ToCoNsri'(TASpiHAi. of Arcs ok Circles UAvixci Radii 1ncreasin(! hy Uniform Increments AB. Produce AB both wayp. With A lis centre and A B as radius, describe a semicircle meeting B A produced inM. With B as centre and Bn as radius, describe the semicircle meeting the line in b. Afrain, wish rmtre .4 iiiri radiH= A l>, describe annthor scmicirclo and .-o ou, alternately using A and B n,s centres, with radii inoreasino at each change of centres, by a distance A B. PUOBLKM 74.-TO Draw a Spiral of Any Ni'.mber of Revolu- tions AND HAVINO a (JiVEN PiTCH. -f. Produce A B and measure off as many parts equal to the pitch A B as the required number of re?olutions (in this case three). Describe circles with centre A and radii AB, A V. Divide the outer circle into any number of equal parts and draw the radii An. A b. etc. Divide A B and B<' each into the same number of equsl parts as the circle. Measure along Ai« a distance from A equal to one of these parts ; along Aba dis- tance equal to two of the piirts, and so on, obtaining the points I,, M, W, etc., in the required spiral. For the second revolution meiisure along A a from A a distanoeequal to A B and one of the parts, along A b, a distance eqoal to A B and two of the parts, and so on, obtaining points in the second revolution in IJ, M', Mi, etc The third revo'ution l^ M-'. 5f-, etc., will be obtained in a similar manner. OF Revolu- the required idii An, Ab, irole. Dg A b a dis- nts I,, n, N, A B and one liart^i and so PHOBLKM 75.-T.. D.AW a I,„,.h,,k Shhai. ck i,s(, Vou.u.r. a HKAjn^HAPKH CAM, Sn,.,,,,,.: ,.„,. Comm. .^u■Ar,v. I •^,.,;KM Assume the axis of r. tnli.m A, the ,„ini„unn an,l .uuximum radii A R and A V l)e.-enl)c nrcles witli cenlre A iii.d riidii A «', A II J.ivide '';-;';^^J '=;--"^|'^rc,,oe c.u.lly in tl.e ..oi,',.. f, k, h. etc.. and draw .!,., radii BhiJe !«<■ i„l„ tl... .;,mo number of ,.,„u.l pirtsastho semicircle in a.b.c etc- Irom the centre A wuh -Jii A» Vb, etc., de.eril.e arcs ,., taeet tl.e r'a i A f. Join tlie points freehand lo form the ram II X <| s. re-imrea curve. PUOllLEM 7«.-To n,;snt„„, a Vnnnos ok t.i,.; Involute or a (ilVK.S ('ll{( I.K IJCI). l.ivi,lc "- >ireumterence into „n,v n„n>I,er of c-,u„I p.rts in the points ». b, c, etc. M.ke no o„uul „. ,h. len.th „f tl.e ,.re a B. bre„u„l to the „rc Bb. e„u„l to twice no, nnd similarly ohtain tl,o ,»,i„„ r. h. j. u, «;c.. incrcsiuK the tangent in e.ich oiise liy a lensth ei|n,il to the reetilie.l are nB ..o a, to e.|ual the lengtli of the eorrc.«|iondinB are. N0TK:-Thi. curve is formed by a ,m,„t in a tine .l,r,..,l when kept tan.ht nnd unwound from a crcuhir dise. tt Is >nueh used in nu'ehanlsm and is one of the forms given lo ilie teeth of wheels. PK<)BM;>r 77. T,) DiiAWT,,,,: Cv, i.,„n (;ivsi.:uATi:h iiv a Point t- IV Till.: CiHt iMiKHiNi i: oi- A (livDN CiH.r.i: lie I). Draw the .li.uneter <'AB and through A and B draw lines A K. B F, at right angles to H<'. Divide the circle into any number of e(|.ual parts in the t>oints n, b, etc. Mark ofl- on A K the points I, m, etc.. at intervals r,,ual to tlie rectified length of the arc Bn. Draw circles witli radii equal to A <' from the centres I, in, ii. etc. Through n. b. I». r, ,1. draw lines parallel to A K. meeting the circles in the points I., M, ST, etc., which will he in the required cycloidal curve. NoTBi-This construction Is equivalent to rolling the given circle along a straight line ; ttie centre of the circle successively .iccupying the points I, iii. n. etc wni e the point <' in the cireiimferei,ce lakes the positions I,. M, N. etc A cycloid IS described by any point in the eireumference of a ,.irelc which rolls on a straight line called tlie dirortor. 1. l.L... I'Uoni.KM 7H.-T., Dhaw thk Ei'i<;Y(|.oii. Describki) by a Point in rni; Circi'mferkncio of a Givkn Gk.vkraitno CiRd.i: B I), Roi.i.iNd o.v A GiVKx DiREcriNii CiRcLi: BE. Join the centres AC. Divide the circle BD into the equal parts B«, »b, etc. With A lis centre ami A «' a« railius, describe an are of a circle l"r. With the same centre and radii A a, A h, A c. A .1 and A 8, describe ares of circles. Divide the circumference B K into e.,ual parts Bf, Tg. etc., having the same length as the ares B,.,»b. be. etc. (non rolling would come into contact with f, b with ir, c with h. etc.) From A ,lraw radii through f. k, h. etc., to meet the are V r in the point- m, a, o, etc. With centre in and radius <'D. describe an arc ofacirde cutting the arc described with the ra lius Aoin I,. This arc I- f represents one position of the generating circle and the point I, the position which the generating point » would then have reached. Similarly with centres ... «, etc., describe arcs meeting the arcs described, with radii A«I. A c, etc., respectively in M, X, etc. The generating point will successively occupy the positions », I„ M, ». «». I*, H. on the curve. The remaining half of the loop is to be similarly formed. PR Witl Divic Take Throi Join t PRC From Then PBOBIiEM 79.-T(> Draw thk Hvpocyloid or Interna EPICYCI.OII) DESCRIBEn BY THE PoiNT D IN THE GeN ERATINd CllUI.E 1)B, KOI.I.INC WITHIN THE DlHE( TIN( ClR( I.E EB. The description for the external epicycloid will serve also for this curve. When the diameter of the generating cirje equals the radius of the direct ing circle, the curve becomes a straight line. k-f--:V;\ 4 •.,\ I PROBLKM «o.-To Trace thk Path ok a I'oim P WITHIN THE Circumference ok a Circi.e a I) WHEN THE CmcUMFEKENCE Roi.l.S ON A HTUAKillT Line AB. (Such curve.s are culled trochoids or cycolids.) With E ns centre and E F as radius, describe a olrclo. Divide llie outor circle into any number of ec|ual jjarts aii'i join these to the centre, cutting the smaller oirole in the iiointa b, c, rolnt« trochoid. SUf tfllOM THOCMOIO PROBLEM «1.— To Draw Epithochoidai. CtrRVEs. These are described when a circle rolls upon a circle and the generating lioint is within or without the circumference of the rolling circle. The former is called the Inferior and the latter, the nuperlor epi. trochoid. The construction is similar to that for the trochoidal curves. I. M N <> i»8 is a portion of the superior epitrochoidal curve traced by the point I. when the circle A D rolls on tho external circumference BAC, and TUVWYZisa portion of the inferior epitro- choidal curve formed by Ihe internal point T under the same circumstances. PItonijKM H2— To Dhaw tim; Hvi'ontoi iioidai, Ci-kvi.s. Theseareiie?crioeilwlien the gciienitiiig circle rolls within the directing circle. The construction is similiirto the foregoing. To find the curve trncej by the point I> when the circle A K tolls within the circumference B A € : Divide the circumference of the circle A K into iiny number of equal parts, and on B A V mark off on either side of A parts equal to these. Join the points so foumi to the centre F outtinB the arc described through the centre H of the rolling circle in n, h, v, <■,•>, KT, etc. Through ihe points in the circumference of the circle having a raclius II B, describe aros from the centre F and from the centres a, b, c, etc., describe arcs with radius e.iuiil to H I> meeting the former in 1., N, <>, etc. This curve, the superior hypotrochoid, will form loops as N O P K. The inferior hypotrochoidal curve, trajc.l by the point T i.f which T IJ V W X is a ?iart„ is described in a similar mam er. I-WOLfTLS I.N Gi:.\KHAI,, As IXVOI.l TICS OF AN El.l.Il'SK, CV(l,(lll), Etc. The path of n point in a flexible inextensiblo thread made to unwind itself ta.igen- tially from a curve, in the plane of the curve, is an Involute. The curve from which the line is unwound is called the evoliito of the involute curve. The involute curve being given, its evolute may, by a converse operation, bo determined. To form an involute, tangents are drawn from the evolute, and the points of tangenoy are ths instantaneous centres of curvature. If from the involute normals be drawn and the centres of curvature obtained, ilie locus of these centres will be the evolute. (See figure and also I'roblem 76.) To obtain (he cen(res of curvature when Ihe character of the curve is not known, it is necessary to assume three points near to each other, and obtain the centre of the circle passing through them. (Prob. 3fi.) As any curvix maybe considered an evolute from which to obtain an involute, so also it may be considered an involute from which the corresponding evolute may be obtained. Cycloids ix General. A Cycloid is described by any point in a curve when the curve rolls on any other curve. .Vs for e.xamjile, an ellipse rolling upon an ellipse. As an exercise, describe the cycloid generated by a point in an ellipse having axes of 3" and 1" when the director is a straight line. L Cl'kvks. ilireoting circle, he curve trnced I circiiuifereiice nto any number ide of A piirt? •e F cutting the rcle in n, l>, v, a railiu; If !>. ' n, l>, r« etc., iirmer in I/, W, 1 form loops as r which T II V PROBliEM «3 —To Dh.vw tiii: Co\( koid Cuhvks of which thk Sthaicht I,i\i: A It is nii; Asvmi'Toik, V l> the DiAMin'EK, AND K TMK Por.i:. -5— i-— M F The line f C is perpcndiculur to A IS. On each side of D mark any points a. b, c. etc, and from F draw straight lines throuijh these points. Make D E eoual to V I», ar.d from the points n, b, o, etc , measure lengths i» l.» b M, etc., « I..' b M , etc., each equal to <' ». The poii;'» «', I., M. arc in the superior Conchoid, and K, I/, M , etc. in the liifvrlor C'Onctaoid. A freehand curve is to he drawn through the points. PKOBLEM 84.— To Dkaw thi: Cissoid Critvi: from thi: Civkx Cikcm;, E C. t s -TO n. m t C At <' draw the tangent line A «'. From K. the opposite extremity of the diameter through the point <', draw any lines meeting the straight line A «' in the point") I. m, n, cte.,aiui cutting the circle in h, k, f, etc. Make A I' equal to R E, t T equal to b E. etc.: o O equal to e !".. n ST cqua! to f E. etc. The points U and T. O and S are in the required curve. Obtain the points S, H, and M, in a similar manner and dr(\w the curve freehand.