Class / .g J 3 Book._ Copyright ]^°_ COPYRIGHT DEPOSnV i'-'W Constructive Drawing. A TEXT-BOOK FOR HOME INSTRUCTION, HIGH SCHOOLS, MANUAL TRAINING HIGH SCHOOLS, TECHNICAL SCHOOLS AND UNIVERSITIES. ARRANGED AND PUBLISHED BY HERMAN HANSTEIN Instructor of Drawing A. G.Lane Manual Training High School, Director of Art and Technical Drawing: Departments. Chicago Mechanics Institute, Chicago Business College, formerly Supervisor of Art and Constructive Drawing Departments Chicago High Schools. KEUPPEL & ESSER CO. THIRD EDITION. GEOMETRICAL CONSTRUCTIONS. CHICAGO 1908 NEW YORK CHICAGO ST. LOUIS SAN FRANCISCO AUG t7,^a08 OUA$S (X^ _JJ^ XKu I »4- i^ s i, COPY B. ^V W^ oa,.2^^^ ENTERED ACCORDING TO ACT OF CONGRESS, TN THE YEAR 1906 3Y HERMAN HANSTEI IN THE OFFICE OF THE LIBRARIAN OF CONGRESS, AT WASHINGTON, PREFACE TO THIRD EDITION A T the request of former assistants and pupils I have compiled and ■'- *■ revised these problems, as a help and for home instruction. This work represents the first year's course, that has been followed the past twenty-five years in the Chicago High Schools and in the Drawing Depart- ments of the Chicago Mechanics Institute. A practical experience of seventeen years in office and shop and his occupation as teacher during the past thirty years have given the author such experience and judgment as to select only such problems as are of practical importance to those who follow architectural, mechanical and engineering vocations, as well as problems which are indispensible to manufacturing and industrial pursuits. The author feels very grateful for the manner in which his former editions were received and hopes this revised third edition will meet with increased favor. HERMAN HANSTEIN, Cliicago, III., July 1908. 361 Mohawk Street INDEX. Tools, implements and their applications Figs. 1 to 7, Plate 1 Alphabets Plate 2 Block Alphabets Plate 3 Bet squares and their applications Figs. 1 to 4, Plate 3 Comers for borders Fig. 5, Plate S DEPISITIONS. Lines and angles Text opposite Plate 2 Planes and surfaces " '" Plate 3 Polygons " " Plate 3 Circle " " Plate 3 OONSTROOTION OP PERPBNDICnLARS. To construct perpendiculars Nos. 1 to 6, Figs. 1 to 6. Plate i To construct perpendiculars Nos. 7 and 3, Figs. 1 and 2, Plate 6 DIVISIONS OF LINES. To divide lines Nos. 8 to 12, Figs. 3 to 6, Plate 6 SOLUTIONS OF ANGLES. To construct and divide angles Nos. 13 to 18, Figs. I to 4, Plate 6 SOLUTIONS OF TRIANGLES. To construct triangles Nos. 17 to 20, Figs. B and 6, Plate 6 and Figs. 1 and 2, Plate 7 PROPORTIONAL LINES. To construct proportional lines Nos. 21 to 84, Figs. 3 to 6, Plate 7 POLYGONS. To construct regular polygons on given bases Noe. 25 to 39. Figs. 1 to 4, Plate 8 To construct equivalent polygons Nos. 33 to 40, Figs. 6 and 6, Plate Sand Figs. 1 to 6, Plate 9 TRANSFERRING POLYGONS. To construct polygons equal to a given one Nos. 41 to 46, Figs. 1 to 6. Plate 10 and Figs. 1 to 6, Plate 11 REDUCTION OK ENLARGEMENT OF POLYGONS. To construct polygons proportional in area or outline and similar to given ones Nos. 47 to 52, Figs. 1 to 11, Plate 12 and Figs. 1 to 5, Plate 13 SCALES. To construct scales Nos. 53 to 56, Figs. 6 au J 6 A, Plate 13 and Figa. 1 to B, Plate 14 CIKCLEa. To divide circles Nos. 57 to 60, Figs. 1 to 5, Plate 15 RECTIFICATION OF ARCS. To rectify arcs Nos. 61 and 62, Fig. 6, Plate 15 and Fig. 1, Plate 16 INDEX.— Continued TANGENTS. To construct tangents Nos. 63 to 66, Figs. 2 to 6, Plate 16 TANGENTIAL CIRCLES. To construct tangential circles Nos. 67 to 81, Figs. 1 to 6, Plate 17, Fi^s. 1 to 6, Plate 18, Figs. 1 to 4, Plate 19 GOTHIC AND PERSIAN ARCHES. To construct Gothic and Persian arches Nos. 82 to 87, Figi* 6 and 6, Plate 19, Figs. 1 to 4, Plate 20 KGG-LINES. To construct egg-lines Nos. 88 and 89, Figs. 6 and 6, Plate 80 OVALS. To construct ovals Nos. 90 to 93, Figs. 1 to 4, Plate 21 ARCHES. To construct arches Nos. 94 to 98, Figs. B and 6, Plate 21, Figs. 1,2 and 4, Plate 22 ASCENDING ARCHES. To construct ascending arches Nos. 99 to 102, Figs. 3 and 6, Plate 22, Figs. 2 and 3, Plate 23 SPIRALS. To construct spirals Nos. 103 to 106, Fig. 1, Plate 23, Figs. 1 to 3, Plate 24 CAM LINES OR ARCHIMEDEAN SPIRALS. To construct cams Nos. 107 to 110, Figs. 1 to 4, Plate 25 CONIC SECTIONS— ELLIPSE. To construct ellipses Nos. Ill to 120, Figs. 6 and 6, Plate 25, Figs. 1 to 8, Plate 26 PARABOLA, To construct parabolas Nos. 121 to 126, Figs. 1 to 6, Plate 27 HYPERBOLA. To construct hyperbolas Nos. 127 to 129, Figs. 1 to3, Plate 28 GEAR LINES— CYCLOID AND EVOLUTB. To construct an evolute No 130, Fig. 4, Plate 28 To construct a cyloid Nos. 131 to 133, Figs, land 2, Plate 29 To construct an epicycloid and hypocycloid (epitrochoid and hypotroohoid) Nos. 184 to 138, Figs. 1 to 3, Plate 30, Figs. 1 and 2, Plate 31 APPLIED CONSTRUCTIONS IN ARCHITECTURE AND MECHANICS. To construct an ornamented Gothic arch No. 139, Fig. 1, Plate 32 To construct a pair of "spur" whet'ls No. 140, Fig. 1, Plate 83 Necessary Tools, Implements and Their Application. Fig. 1, Plate!.— .4. drawing-board, made of well- seasoned white pine, poplar (whitewood) or basswood, the lightest of our woods, answers this purpose best, as these woods are evenly grained and do not offer great obstruction to thumb-tacks, by which the drawing-paper is fastened to the board. The under surface of this board should be provided with two parallel dovetailed grooves, 3 or 4 inches from edges O and O' and right- angled to the grain of the wood, to receive not too tightly fitting cleats, at which the board may shrink, to prevent its splitting. The cleats therefore should not be glued in the grooves to receive them. When one draws with the right hand, the straightedge, called T square (T), and triangle S, called set square, are operated with the left hand, and when one draws with the left hand the set and T square are operated with the right hand. The T is used only on one edge of the hoard. Pigs. 1 and 2, Plate 3.— Set squares (Triangles). — One set square of 30" and GO" and one of 45" (degrees) are required, as shown in Plate 3 and these should be tested for accuracy before ad- mitted to practical use. I'est. — Place the set square with one right- angle side to the T, as shown in Fig. 1, and draw with a hard (4H) well-pointed lead pencil a line on side a b. Reverse the set square on a b as an axis, and if the line drawn and the side of the set square coincide (fall into one) the angle is a right angle, while a convergence will show twice the angle to be corrected, and such a set square should not be used until it is made true. Likewise investigate the T square before using it. Figs. 3 and 4. — Plate 3.— Figs. 3 and 4 show the different angles possible to be drawn by means of both set squares and the T. Fig. 2, Plate 1. — TTie protractor is a semi-circular instrument made of brass or celluloid, for the purpose of measuring the size of an angle. Point C represents the center of the semi- circle, which is divided by radii into 180 equal parts, termed degrees. In measuring an angle, such as the angle B C A, place the in- strument with its center at the vertex (the inter- section of the sides of the angle), and one side to coincide with the diameter of the instrument. Note the number of divisions on the intervening arc, which is 137 (read 137° degrees), which re- presents the size of the angle. A degree is fur- ther subdivided into 60 parts called minutes ( '), which are in turn subdivided into 60 parts called seconds ("), THE SET OF DRAWING INSTRUMENTS. Necessary Tools, Implements and Their application. — Continued. sufficiently to have both blades touch paper equally to allow an even flow of the ink. The leg which carries the center of the compasses should have a vertical position so as to avoid a tapering of the hole in the paper by its revo- lution. Pigs. 4 to 7, Plate l.—The very best is none too good. A set should contain one pair of compasses, Fig. 4, with needle-point center. Pig. 4 D, a lead pencil attachment, Pig. 4 B, a ruling-pen for circles. Pig. 4 A, one pair of dividers. Pig. 5, and one or two straight ruling-pens. Pig. 6, of different sizes. Por boilermakers, machinists, architectural iron constructors, etc., a set of bow instruments is a valuable addition to the above. Pigs. 7 and 7 a, Plate l.—T/ie lead (6 H) for the compasses is bought in sticks of 5 in. in length and ^ in. thick. Take a length i in. longer than the length of the hole in the attachment to receive it. Give the lead the shape shown in Pig. 7, which is most conveniently done on a piece of emery paper or a fine file ; then take off the corners as indicated by the lines G K and H I Pig. 7 A. Insert it with a flat side towards the center of the compasses and clamp it tight with the clampscrew S, Pig. 4 B. The main joint near the handle ought to move with ease, and one hand should be suffi- cient to open or close dividers or compasses easily. The straight pen and the pen for circular ruling must be treated most carefully. Their blades are of the same length, not so pointed and sharp as to cut the paper, and when filled with ink should be entirely free of ink on the outside. In inking circles, the legs of the compasses should be bent at the joints P and O (Fig. 4) The correct position of the compasses is shown in Pig. 4, where the line M N represents the surface of the drawing-paper. A convenient arrangement to keep the plates of the course for later reference is used in the Chicago High Schools. The sheets of paper of 1 1 in. X 17 in. are perforated and seamed by laces in a porte- folio of 12 in. X 18 in. A rectangle as a bor- der line of 10 in. X 15 in. encloses the drawing surface which is divided into six equal squares 5 in. on a side, each to receive one construc- tion. See Plate 4. Larger spaces however should be used to execute accurately some of the constructions, for which the proportional sizes may be ascertained from the correspond- ing plates. Draw the lines light and carefully with Dixon's V H (very hard), Paber 4 H (Sibe- rian), or a Hartmuth 6 H (compressed lead) pencil, having a fine round point. Inking the drawing.— 'Execnte all construc- tions in pencil, to admit of corrections, when necessary, before inking them. It is also ad- visable for the inexperienced to write the re- quired text on the drawing in pencil, to distri- bute letters and words regularly in the avail- able space beneath each drawing, as shown in Fig. 1, Plate 4, before writing with Indian ink. Plate i. Hnimtcln's Comtnictivc Diawlng. Plates 2 and 3.— AlpTiabeta.— Several styles of lettering tor titles of drawings commonly used, are shown on Plates 3 and 3. Plate 8 —In Fig. 5, ABCDEFQ and H sliow a few samples of corners in border lines for elaborate work. The following distinctions of inked lines in drawing are made to recognize readily all that pertains to problem^ construction and reiult. The problem line is drawn/ine and uninterrupted. The construction line isfl^icand dashed. The result, a strong, uninterrupted line. Begin inking with construction arcs and circles, then the cir- cular problem lines, and then the circular result lines. This is done so as to save time, to avoid the change of tool in hand, and not to clean and re-set the pen oftener than necessary. Construction straight lines are drawn next very fine and dashed, corresponding to construction arcs and circles, and lastly the Result straight line, to correspond to result arcs and circles. The inking of a drawing is a recapitulation of each construc- tion, and this important work should be executed with great care. A. postulate is a statement that something can be done, and is so evidently true as to require no reasoning to show that it can be done. An a^^iom is a truth gained by experience, and requiring no logical demonstration. A theorem is a truth requiring demonstration. LINES AND ANGLES. A right Une is the shortest distance between two points. When the term line alone is used, it indicates a right line. A vertical line is the "plumb-line " ; a horizontal Une, one making a right angle with the vertical and a line of any other direction, is called oblique. A curved line or curve changes its direction in every point. When two lines lying in one plane, on being produced in either direction, do not intersect, these lines are said to be parallel. Two lines that intersect, or may be made to intersect, are said to form an angle. The point of intersection of these two lines, called sides, is the vertex of the angle. When two such lines intersect each other, so that all four angles formed are equal, we say they are right angles. The common vertex, of these four right angles may be assumed to be the center of a circle, which by diameters is divided into 360 equal parts, called degrees ("). Each angle contains % of 360" = 90°, which is the right angle. An angle greater than 90° is an obtxise angle; an angle smaller than 90° is an cicute angle. When the two sides of an angle form a straight line, the angle is called a straight migle, and its magnitude is 180°. Generally we designate an angle by three letters, for instance, . b a c or c a b ; then the middle letter (a) indicates the vertex, while the sides are b a and c a. Plate 2 Engineering 'Sct^ipt, (Roman) 123+5 (vbcclefghi.jkl7n nopqrstuvnvxi^z. ejsoo. ABGDEFGHIJh LM^OPQR SnrVWXYZ. AF\C^iTtQTVF\AL. l2_5-=f^ abcdefgf\ijKlTT\T\opcfKsluVW\ijz. 6/590. Shop Skelston. abcdeF^hijklmnopciH-stuvwxyz. 12545 ABCDEFGHIJKLM NOPQRSTUVWXYZ. 87390. ABCDEFGHU KLIVINOPQ RSTUVW>{yZ ipc|: Ha».«(ci»'9 Conslnictlos Drauiinff. PLANES AND SURFACES. A plane lias two dimensions— length and breadth. A surface is the boundary of a body. Surfaces bounded by right lines are called polygont. Begular polygons have equal sides and equal angles ; they are equilateral and equiangular. POLYGONS are: The triangle, " tetragon o '* pentagon, which has 3 sides, quadrilateral. heptagon, octagon, enneagon or nonagon. undeoagon, dodecagon. 12 etc. The triangles are: The equilateral triangle which is also equiangular; the ieusecles triangle, having two sides equal, and the scalene triangle, whose sides are unequal. An obtuse and a right-angled triangle have one obtuse and one right angle respectively. An acute angled triangle has three acute angles. The side or "leg" opposite the right angle in a right-angled triangle is called the hypotenuse, the sides or legs forming the right angle are the catheti. The sum of the squares constructed on the catheti is equivalent to the square erected on the hypotenuse. The sum of all angles in a triangle is equal to two right A line drawn from a vertex of a triangle perpendicular to the opposite or produced opposite side is called its altitude or height. QUADKILATEKALS. The regular quadrilateral is the square. (See definition of regular polygon.) Lines joining the middle of the opposite sides are called diameters. Lines joining opposite comers are called diagonals. In a square the diagonals are equal. A quadrilateral whose opposite sides are equal and parallel is called a parallelogram, A rectangle is a parallelogram whose angles are right angles and adjacent sides unequal. A rhombus is an equilateral quadrilateral with unequal diagonals and equal opposite angles, A trapezoid has only two parallel sides. A trapez.inn is an entirely irregular quadrilateral. Fig. I. Plate 15.— Definition.— A circle is a portion of a plane bounded by a uniformly curved line, called the circumference, all points of which are equally distant from a fixed point within, called the center. The distance from the center to any point of the circle is called the radius. The connecting line of any two points of the circumference is caUed a chord. If the chord is produced to any point outside the circle, it is called a secant. The chord tiirough the center is called the diameter. Any arbitrary part of the circumference is called an arc. The arc that forms the fourth part of the circumference is called a quadrant; the sixth part a sextant; the eighth part an octant; while half the circum- ference is called a semi-circumference. The area comprised by two radii and the intervening arc is called sector; the area comprised by a chord and the corresponding arc is called a segment of a circle. In Fig. 1, C D, C B and C A are radii, G H is a chord, E I F is a secant, A B is a diameter, G J H an arc, area D C B L D is a sector, area H G J H a segment, tract A J U B a semicircle. Postulate.— T>T3iVr a circle, if the center and the radius are given. Plate 3. BLOCK, EF6HIJKLMNOP0R STUYWXYZ. 1234567890. &. sijjifloiyuj^ji. ®^/i^iJjj;iL>7jj^JDj^i;j-F] rtr: Hmwtein'S'CQi\atntctivc DrawTnQ. CONSTRUCTIONS. 1. — Fig. 1. — Problem. — Tu erect a perpendicular at a given poird in a given line, or to bisect a straight angle. Solution. — Let M N be the given line and A the given footpoint of a perpendicular. From A as a center and with any radius describe the arc B, C; B and C are equidistant from A and are the centers of arcs with equal radii greater than B A, which intersect at point D. Draw the line D A, which is perpendicular to the line M N, in point A. 2. — Fig. 2. — Problem. — To draw from a given point a perpendicular to a given liiie. Solution. — With the given point A as a center describe an arc intersecting the given line M N in two points, B and C. From B and C as centers and with equal radii draw arcs intersecting at D. Con- nect points A aud D by the line A D, which is per- pendicular to M N. 3.— Figs. 3, 4, 5 and 6. — Problem.— To erect a per- pendicular at the end of a given line, M A. SoZirfioft.— Take any point C outside of M A as a center, and with a radius C A describe an arc in- tersecting MA at D. Draw the diameter D C B. A line drawn through B and A is the perpen- dicular to M A. 4. — Fig. 4. — Solution. — From A as a center and any radius describe the arc B N, at which make B C = A B and pass through points B and C the line B C indefinite; make then C D = C B. A line drawn through A and D is the required perpendicular. 5. — Fig. 5. — Solution. — Describe from A as a center and any radius the arc B C E. Make E C = C B = B A, and from E and C as centers and with equal radii draw intersecting arcs at D. A line drawn through D and A is the required perpendicular. 0. — Fig. 6. — SoZtttioji.— From A towards M lay down 5 equal units. With A as a center and 3 units as a radius draw the arc 3 B indefinite, and 4 as center and 5 units as radius cut the arc at B. A line drawn through B A is the required perpendicular. Plate 4. Fig. 1. Fig. 2. Fig. 3. / 1 1 1 B / / / c/ /\ / \ 1 1 1 M 5\:- — ■ -■;> /CH''' \ \ \ \ \ V^^ -' 1 ( 1 A . ; B \ ; ' / \ / \ / > . '' E ^ A" Fig. 4. A C B Fig. 5. B Fig. 6. "^ i ''!; \v 1 \\ \ \ ■w \ N. \ V \ \ .' \ \'q 3 ^^-0^11 « Hanalein's Constructive DraKlng. SOLUTION OP ANGLES. 13, — Fig. 1. — Problem. — To construct an angle equal to a given one. Solution. — Angle C A B is the given angle. When the vertex O and one side O N of the angle to be constructed are given, describe with O as a center and A C as a radius the arc E D, and from D as a center with the radius B C the arc at E ; draw the line E O. Angle C A B = angle E O D. 14.— Fig. 2.— Problem — To bisect an angle. Solution. — Let B A C be the given angle. With A as a center and a radius A B draw the arc B C. B and C are the centers for arcs with equal radii, intersecting at D; draw line from A through D, which divides B A C into two equal parts. 15.— Fig. 3.— Problem.— To trisect a right angle. Solution. — From vertex A, with the radius A B, draw the arc B C. With B as center and the same radius draw the arc A E, and from C the arc A D; draw lines through D A and E A. Angle B A D = D A E = E A C. 16.— Fig. 4.— Problem.— To trisect any angle. Solution.— Ijei C A B be the angle to be trisected. Describe with A as center a semi- circle BCD, which intersects the prolonged side B A of tbe angle at D ; draw from C an arbit- rary line OEM and make E F = E A, and draw F G C; then make G H = G A and draw H I C. An additional operation will not be necessary, as the lines will fall so close together as to almost coin- cide, and it is angle C H B which equals i C A B. This construction is convenient for angles up to 90"; and in case of the trisection of an obtuse angle we bisect first and then trisect, so that the double third of the bisected angle is equal to the third of the given obtuse angle. SOLUTION OP TRIANGLES. 17. — Fig. 5. — Problem. — To consti-uet a triangle when the three sides are given. Solution.— Liines 1, 2 and 3 are the sides given. Lay down line 6 = line 1. From C as center, with line 2 as a radius, draw an arc, and with line 3 as radius and center B another arc, intersecting the first arc at D. Draw lines D C and D B ; then D C B is the required triangle. 18.— Fig. 6.— Problem,— To construct a triangle of which tioo sides and the included angle are given. Solution. — Construct angle D, and from its vertex cut off the sides 1 and 3, that is C B and C E, and draw line E B; then E B C is the required triangle. Fig. 1. 0} Fig. 2. Plate 6 Fig. 3. Fig. 4. Fig 5. Fig. 6. Hanstein's Constructive drawing. 19. — Fig. 1.— Problem. — To construct a triangle of which one side (1) and the two adiacent angles D and E are given. Solution. — Lay off C B equal to line 1; transfer the angles D and E on line C B, and prolong the sides to intersect at P; then triangle C F B is the required triangle. 20.— Fig. 2. — Problem. — To construct a triangle of which one side (1), one adjacent angle D and one opposite angle E are given. Solution. — Construct C B equal line (1) and angle D at C as before; at an arbitrary point M on line CM draw angle C M N = E, and parallel with M N the line B F. F is the third vertex of the required triangle C P B. PROPORTIONAL LINES. 21.— Fig. 3.— Problem.— To construct to three given lines a fourth proportional. Solution. — Let 1, 2, and 3 be the given lines. Lay down an angle M A N of about 40", and from A cut the segments A 1 = line 1, A 2 = line 2, A 3 = line 3; draw line 2 1, and with it parallel the line 3 x. A X is the required line. 1 : 2 = 3 : A x. 22. — Fig. 4. — Problem. — To construct to two given lines a third proportional. Solution. — Let 1 and 2 be the given lines. Lay down the angle as before, and from A cut the seg- ments A 1 = line 1, A 2 = line 2, A 2' = line 2. Draw line 21, and parallel with it 2' x. Ax is the required line. 1 : 2 = 2 : A x. 23. — Pig. 5. — Problem. — To construct a mean pro- portional to two given lines. Solution. — Let 1 and 2 be the given lines. A B -f B C is the sum of the given lines 1 + 2. Find point D, the center of A C, and with a radius D A, draw the semi-circle A X C. Erect at B a perpendicular B X, which is the required line. A B : B X = B X : B C. 24. — Fig. 6. — Problem. — To construct to a given line major and minor extreme proportionals. Solution. — At point B of the given line A B erect a perpendicular B D = i A B, and draw line A D P indefinite; with D as center, D B as radius, describe semi-circle EBP, and from A as center, A E as a radius, draw arc E X. The line A P : A B = A B : A X. Fig. 1. A Fig. 2. Plate. 7. Fig. 3. \ s _-b "s- - Fig. 4. \ \ \ \ \ \ i ^v. 2 I Fig. 6. HarxsUiri's. Constructnvc Drmoing, POLYGONS. 25.— Fig. 1.— Problem.— To construct a reoular triangle on "■ given bane. Solution.— A B is the given base With A and B as centers and A B as radius draw arcs intersecting at C. Draw the lines C A and C B. A C B Is the required regular triangle. 36.— Fig. 1.— Problem.— To eon-strttct a regular hexagon on a given base. Solution.— het A B be the given base. Construct on this a regular (or equilateral) triangle. The vertex C is the center, and C A = C B the radius of a circle, in which a regular hesa- gon A B D E F Q, v;ith A 6 as side, can be inscribed. Corollary. — A regular hexagon may be divided into six equal equilateral triangles, the common vertices of which Ue in the center of it. 37.- Fia. 2.— Problem.— To construct a regular heptagon at a given base. Solution,— Draw with the given base A B the equilateral tri- angle A 6 B. as in the previous construction. From center D of A B draw the line D 6 13 perpendicular to A B. Divide 6 A into six equal parts. These parts transfer on line 6-1.2 and number them, 7, 8, 9, 10, 11 and 13. Point 7 is the center, and 7 A the radius of a circle, in which the regular heptagon A B C D E F G, with A B as aide, can be inscribed. 28.— Fig. 2.— Problem.— To construct a regular polygon loith more than 6 sides. Solution.— With points 7, 8, 9, 10, 11 and 13 as centers, and 7 A, 8 A, 9 A, 10 A, 11 A and 13 A, respectively as radii, draw circles in which the line A B rs repeated chord will form the regular heptagon, octagon, enneagon, decagon, undecagon and dode- cagon. Bemarfc.- Regular polygons with greater number of sides are rarely used in practice, and are therefore omitted here. 29.— Fig. S.— Problem.— To construct a square at a given base. Solution.— \je.t A B be the given base. Draw at A and B per- pendiculars with set and T square, and make A C = A B, and with T square draw CD. A O D B is the required square. 30.— Fig. S —Problem.— To construct a regular octagon at a given base. Solution — In the bisecting point H of the given base A B erect a perpendicular. H F, at which make H E = A H and E F = E A. F Is the center and F A the radius of a circle, in which draw A B eight times, as repeated chord, to complete the re- quired octagon ABGHIJKL. 31.— Fig. 4.— Problem,— To cointrifct a regular pentagon at a given base. Solution.— Let A B be the given base; produce it towards N. Erect at B a perpendicular, B D = A B. Bisect A B by point C; with C as center and C D as radius draw arc D E. With A and B as centers and A E as radius draw arcs to Intersect at F. With F and A as centers draw arcs intersecting at G; and from F and B as centers, with the same radius A B, draw arcs inter- secting at H. Connecting B H, H F, F G and G A by lines we complete the required pentagon A B H F G. 32.— Fig. 4.— Problem.— To constn«;t a regular decagon at a given base. Solution.— Let A B be the given base. Follow the construc- tion of the pentagon until the position of i)oint F is found; this is the center, and F A the radius of the circle, In which as re- peated chord the line A B will complete the required regular decagon ABIJKLMNOP. 33.— Fig. 5.— Problem.- To construct triangles equivalent to a given one. Solution.— Let A B be the given triangle; draw line M N parallel with A B through point C. Locate an arbitrary point E or G in line M N, and draw lines E A and B B, and G A and G B. Triangle A B B = A C B — A G B. It one side of the tri- angle is called the base, a perpendicular drawn from the oppo- site vertex to the base, or produced base, is the altitude or height of the triangle, as E F, C D and G H. Theorem. — Triangles of equal base and altitude are equi- valent. 84.— Fig. 5, A.— Problem.— To construct paraUdograms equiva- lent to a given one. Solution. — Let A B D C be the given parallelogram, with base A B. Draw the line M N parallel with A B, make B F and G H = C D, and draw lines E A, F B, G A and H B. The parallelo- gram BFBA — CDBA = GHBA. In a polygon any right line which passes through two non- consecutive vertices of its circumferential angles is called a diagonal. 27ieorem.— Either diagonal divides the parallelogram into two equal triangles. 3B.— Fig. 6. — Problem.- To construct a rectangle equivalent to a given triangle. Solution.— A B C may be the given triangle, and C F its alti- tude. Bisect C F rightangularly by line D E, and erect the per- pendiculars B £ f nd AD. A D E B is the required rectangle Fig. 1. Plate 8. Fig. 3. Fig 4. Fig. 3. a Hajisleln's Coiistniclfve Drawing. 36.— Fig. 1.— Problem.— To construct a rectangle equivalent to a given trapezoid. Solution. — Let A B C D be the given trapezoid. Bisect rightangularly its altitude LM by the line I K, which bisects also the sides B A and C D in I and K. Perpendicular to I K, through I and K, draw F G and E H to intersect the produced B C in E and F. F E H G is the required rectangle, equivalent to the trapezoid A B C D. 37. — Fig. 3.— Problem.— T/ie side of a square is given: to construct the sides of squares that are twice, three times, four times, etc., as great as the square over the given line. Solution. — Construct a right angle B A 1 ; make B A and A 1 equal to the given side of the square; then lay off successively A2 = B1,A3 = B2, A4 = B 3, etc. A 2, A3, A 4, etc., are the sides of squares that are respectively twice, three times, four times, etc., the area of the square over A 1. 38. — Fig. 3.— Problem.— To construct a triangle equi- valent to a given irregular pentagon. Solution. — Let A B C D E be the irregular penta- gon. By the diagnols A and C E divide it into three triangles A B C, C A E and C D E. Produce the base A E to the left and right indefinitely, and parallel to C A draw the line B F ; connect C with F ; then draw D G parallel with C E and connect C with G. The sum of the triangles C F A + C A E + C E G is equal to the triangle CFG, which is equivalent to the irregular pentagon ABODE. 39. — Fig. 4.— Problem— 2'o co7istruct a square equi- valent to a given triangle. Solution. — Let CFG, Fig. 3, be the given tri- angle. Construct a mean proportional between half the base F G and altitude C H, as shown in Fig. 5, Plate 7, by making I K -^ i F G, and K L = C H. The sum I K + K L is the diameter of the semicircle I N L. Erect at K a perpendicular, which is intersected by the circle in N. N K is the side of the required square, and N O P K is the square, which is equivalent to the triangle CFG and the irregular pentagon A B C D E. 40.— Figs. 5 and 6.— Problem.- To tramform anir- regular heptagon into an equivalent triangle and square. Solution.— Let ABCDEFG be the irregular heptagon. Draw line C A, and parallel to it B N; connect N and C by line N C. Triangle C N A = C B A. Treat the triangle E F G in a similar way, and you have transformed the heptagon into the irregular pentagon N C D E M. Proceed as in Fig. 3, and transform the pentagon into the triangle D H I; transform this into the square P Q R O, Fig. 6, which then is equivalent to the given hep- tagon ABCDEFG. Plate 9. Fig. 1. Fig 2. C f;-. 3. ; 6 L C E B ^ Z' \/' ' ^ \ / " V \ /->.-. -.-.^ \ \ \ A MHO A. 1 ^3 °". M F A H E G Fig. 4. R^'. 5. Fi^-. 6. / / 1 1 1 1 1 N \ \ \ ; # \/A / V / / / 1 1 1 L P Q ! : M A /M^y N \ \ \ K 1 P L H " L M C ' ° i •i; V Wan3(eiri's Ci>ns(nM^(u"c Draiofno. TO TRANSPEK POLYGONS. 41. — Figs. 1 and 2.— Problem.— Tb coiistruct a poly- gon equal to a given one by triangles. Solution. — Let Pig. 1 be the given polygon. Divide it by diagonals into triangles. Draw line A' B' parallel and equal to A B. Upon this con- struct triangle A' B' C equal to triangle ABC. Lay off the remaining triangles of Fig. 1 in the same order and position, starting from side B C; then polygon Pig. 2 is the required one. 42. — Figs. 3 and 4.— Problem.— 2'o construct a poly- gon equal to a given one by sectors. Solution. — Polygon Pig. 3 is given. From center O with any radius describe circle C B D. Draw from center O a radius to each vertex of the poly- gon to intersect with the circle. Locate center O', Fig. 4, and with radius O' D' equal O D describe the circle D' B' C, and draw O' D' parallel to OD; make arcs D' B' = D B, B' C = B C, etc., and pass lines through points D', A', C, etc.; further make O' E' = O E, O' A' = O A, O' F' = O F, etc., and by connecting points E' A' F', etc., complete the required polygon. 43. — Figs. 5 and 6. — Problem. — To construct a poly- gon equal to a given one, by co-ordinates. liemark. — In the plane of drawing a convenient line is drawn (horizontal), called the axis of ab- scissae; the position of the diiJerent vertices of the given figure is determined by perpendiculars (or- dinates), from these vertices to the axis of abscissae. Take any convenient point, A, on this axis and draw a perpendicular to it, M N. This line is called the axis of ordinates, and reckoned from this point A (called the origin) the segments deter- mined by the foot-points of the ordinates are called abscissae. Abscissae and ordinates together are called co-ordinates. Solution. — Fig. 5 is the given polygon. Through any vertex (origin) draw a horizontal, A R, then M N becomes the axis of abscissae. Draw the ordinates from each vertex or principal point for transmission perpendicular to A R, the axis of ordinates. Next draw A R', Pig. 6, and lay oflf A B, AC, AD, etc., = A B, A C, A D, etc., of Fig. 5. Erect the perpendiculars A A', B B', C C, C", C", etc., and make A A', B B', C C, O C", C C", etc., equal to the corresponding perpendi- culars in Fig. 5. Connect A and A', A' and B', describe with radius C'C, center C, arc B' C", etc., and complete the required polygon, Pig. 6. Fig. 1. Fig. 2. Plate 10. Fiff. S. --^A« Fig. 3. Fig. 6. Han.ftein^i Conftnicttvc Drawing. 44.— Figs. 1 and 2.— Problem — To construct a poly- gon equal to a given one, radAating in a circle. Solution. — Let A E D G, etc., be the given poly- gon. Describe withi AD, A E, etc., as radii and A as center tlie circles C D, E F G, etc., and make P' E' = F E, F' G' = P G, etc. Connect D' and E', D' and G', etc., and complete the required polygon. Fig. 2 shows the construction applied to other polygons. EemarJc. — This construction is used conveniently to draw a rosette in which an ornamental unit occupies a sector division of a circle. 45.— Figs. 3 and 4.— Problem — To coiistruct sym- metric polygons or outlines. Solution. — LetL M, etc., be the given outline as a profile of the base of a column. Draw the hori- zontals L L', M M', etc., and the axis of symmetry R N. Make A L' = A L, B M' = B M, D O' = D O. etc. Connect L' and M', etc., and complete the re- quired symmetric profile of the base of the column. 46. — Figs. 5 and 6. — Problem.— To construct an ir- regular outline equal to a given oiw. Solution. — Let B A C be the given outline. Cover this with a series of equal small squares and con- struct in Fig. 6 the same number of equal squares arranged as in Fig. 5, and transfer the points of intersections of the irregular outline with the sides of the squares ; make M' A' = M A of Fig 5, and M' B' = M B, etc. Connect B' A' C by a free-hand line and complete the required irregular outline, Fig. 6. Fig. 1. Fig. 2. Plate 11. Fig. 5. I [__!_ Fig. 3. Fig. 4. Hanstein's Constructive Drawing. TO SEDUCE OB ENLARGE POLYGONS IN ODTLIHE OB AREA. 47.— Fias. 1, 2 and 3.— Problem.— To construct a polygon efm ilar to a given one of 4 its circumference. Solution.— Let D A B C, etc., Fig. 1, be the given polygon. Construct the Scale Fig. 2. A perpendicular O 7, longer than the longest side of the given polygon, is divided into 7 equal parts ; draw a horizontal line O N of an arbitrary length and connect points 7 and 4 with N by the lines 7 N and 4 N. O 4 is A, — 4 7 is a of the line 7. All lines between O N and 7 N and parallel to O 7 are divided by 4 Sf and 7 N in the same pro- portion. To obtain the length of A' B', Fig. 8, place line A B in the scale as indicated by line A B' B, of which A B' is ^ of line A B. Transfer the remaining sldei of the polygon by parallels and find of each the proportionate length in the scale Fig. 2, as shown by line A B; D' A' B' C, etc., is the required polygon. 48.— FiQS. 1, 4 and 5.— Problem.— To construct a polygon Sim ilar to a given one, having * its area. SoJution.— Let D ABC, etc., Fig. 1, be the given polygon. On a horizontal line O 4 lay down a division of 7-{-.4 squal parts and malce O 4 the diameter of a serai-circle O M 4. Erect at point 7 the perpendicular 7 M and draw lines M O and M 4. Then make line M B' equal to A B of the given polygon and draw B' B" parallel to O 4; A" B" (Fig. 5) — M B" in the scale Fig. 4. In relation to the side A B of the given polygpn. A" B" is the side of a polygon, whose area is } of the given one. Treat the re- maining sides of the polygon similar to the side A B and com- plete the required polygon D" A" B" C", etc. 49.— Fig. 6.— Problem.— To construct simitar polygons which have I the circumference and | the area of a given one. Solution for circumference reduction.— Let D C B A E, etc. , be the given polygon. From any point O therein draw radii to the vertices D C B A E, etc., and divide any one radius (O D) into 5 equal parts. Parallel to D C from point 3 draw D'C, with B, 0' B', etc., and D' C B' A B, etc., is the required polygon. Solution for area reduction.— Hake radius O D the diameter of the semi-circle O N D and erect at division point 3 the per- pendicular 3 N and draw N O. Make O D" =« O N and proceed as before in drawing D" C" parallel with D C, C" B" with C B, etc. D" C" B" A" E", etc., is the required polygon. 50 — Figs. 7, 8and9.— Problem.— To reduce any irregular out- line in proportion 8 :5. Solution.— Let Q H I K be the given irregular outline . Cover the given outline by a net of equal squares, the sides of which we reduce by the scale, Fig. 8, to A' B' = % of A B. Draw with A' B' as unit the same number of squares as in Fig. 7. Transfer the points of intersection of the irregular outline with the sides of the squares, in reducing their distances from the vertices of the squares by scale Fig. 8, and transfer into Fig. 9. Connect these points by a free-hand line, which is the required reduction of the irregular outUne. Treat the surface reduction, Fig. 11, with the assistance of the scale Fig. 10 in a similar way, and we obtain the reduction in area. Plate 12. Fig.l FTg.3. Fig. 5. Fig. 4. Fig. 6. Fig. Fig.ll. Fig. S. Fi ?■ 9 r ■^ ^'f ■( 6' \ 1 V-41 1 /■n-E/ jAE" "^ ^'i' / I \ -4 FF 1 ^~ HffTMfcfn's Cmr.'rtrHoMrj'- Dtaieinfc 51.— Figs. 1, 2 and 3.— Problem.— ro construct a polygon siinilar to a given one, and of | its cir- cumference. ( Transfer hy triangles. ) Solution. — Let A B D C, etc., be the given poly- gon. Construct the linear scale in proportion 2:3 Fig. 2 similar to Fig. 2, Plate 12, and divide the given polygon by diagonals into triangles. Line A B' in the scale (Fig. 2) = A' B' of the polygon Fig. 3, whose circumference contains 3 units to 2 of the given polygon. Transfer and complete by tri- angles the required polygon A' B' D' C, etc., Fig. 3. 53.— Figs, l, 4 and 5.— Problem.— To construct a polygon similar to a given oiie, which contains S to each S square units of the given polygon. Solution. — In the scale Fig. 4 the diameter of the semicircle consists of 3 + 3 equal parts; erect 2M. Draw M O and M 3. Make M B' Fig. 4 = A B of the given polygon and draw B' B" parallel to O 3, A" B" = M B" of the polygon. Fig. 5, whose area has 3 square units to 2 of the given polygon. Transfer and complete by triangles the required polygon A" B" D" C", etc., Fig. 5. 53.— Fig. 6.— Problem.— To construct a scale of da cimal division. Remark. — Small subdivisions of a unit which we cannot accurately perform with the dividers are constructed in Figs. 6 and 6 A. Solution. — Let line R 7=8 centimeters, RO=RF = 1 cm. The decimal subdivision (millimeter, mm) is obtained by dividing O N into 5 equal parts by the horizontals in points A, B, c and d. Bisect F N and draw lines 5 R and 5 O; line a 1= I'o, B 2 = i?o, c 3 = i'^,, etc., of O R, or 1, 2, 3, etc., mm. the required division. 54.— Fig. 6 A.— Problem.— To divide a centimeter into 100 equal parts. Solution. — Upon a straight line A 7 lay off eight units (cm) and construct squares on these distances. Let first square be A B N O. Divide sides A B and B N into ten equal parts (mm). Draw horizontals through division points on A B. R, being the first point of division from N to B, is connected with O, and through the other points parallels to R O are drawn between B N and A O. These parallels subdivide the millimeter (mm) into tenths- Plate 13. Fig. 1. Fig. 2. Fig. 3. Fig. 4. STJj^t^X . 6A4 V .\ -M-A- 6 f — H ' Fig. 6. A {^ai^tctioii g^xj,f« in 31tx:t*x HniKsfein's Constnictirc DraniC(7. 55. — Fig. 1. — Problem. — To construct a scale in which an inch is divided into 6Iiihs. Solution. — Let A 2= 3 inches. Divide BN and B A into 8 equal parts each and complete the scale in the manner explained in Problem 54, Plate 13, Fig. 6 A. Line R O divides line R N = i in. into 8 equal parts, hence into 64ths. Example : Take from this scale a line of 1|J inch ( |J = l + gV). From O to 4 = f in.; follow the oblique line upward to the 5th horizontal point, N. Line N A = I in., A B = /:[ in. and B M = 1 inch and line N M = IJI in., as required. REDUCTION SCALES. 56.— Figs. 2 and 8.— Problem To construct a de- cimal rediiction scale and draw by co-oi'dinates a polygon whose equations are indicated at tables A and B, Fig. 2 A. Remark. — To draw the scale and polygon in con- venient proportion let the unit O A = 2i in., which may represent 100 feet. Solution. — Let O A be the unit to represent 100 ft. in the decimal reduction scale and let A 300 = 3 such units. Divide O A, A B and B N into 10 equal parts, draw horizontals from 9, 8, 7, etc., and the oblique parallels with R O from division points 10, 20, 30, etc., and we have the required decimal scale. Example: Take from this scale a line to represent 178 feet. Begin at point O, pass to the left to 70, then upward the oblique line to the third horizontal point R. Line r a = 70 ft. a b = 3 ft. and B S = 100 ft., and ra + ab + bs = 173 feet. The polygon, Fig. 3, is constructed with this scale. Bemark.—lt the scale. Fig. 2, is used as a reduc- tion scale in which O A represents 1 ft. , we shall have to divide O A into 12 equal parts (inches), etc., and the scale will represent j', of actual dimension. Plate 14. Fig. 1. SucA -bivliei) i,n,64'-" Fig. 2. A Sci^e^ A A B A C A D A E A F AG A H A I AJ A K _ A L = 5. Ft 27 ■ 4 8,5- 5 8. » 1 5.5" 10 9.' II?.- 128 " 13 9. » I 52 I 5 3. %^U^ B A M = 2 2 K BR „ 3 6 5.Ft C — 51. » uiy as 7 5 " V v _ 2'5 5 " PF' 3 2 (iCV „ 6 3. " HH' -_ 5 0.5 .' 1 r _ 32 » ^J J :^ 50.5 " KK' _ 3 1.5 • 1 1 ' 109.5 • MM' = 9 • Fig. 2. Sl/abu 1 / \ Fig. Plate 21. Fig. 3. Fig. 5. Hanstcln's Conetnictivc Drolving. 96.— Fig. I.— Solution 3.— Let A B be the span, and C D the altitude. Construct with A C the equi- lateral triangle A E C, and make C F = C D, and draw D F G. Parallel with E C dr aw G H I ; points H and K are the centers, A H the radius to arcs A G and J B, and I the center, I G the radius to arcs G D J. Proceed as in Sol. 2, and complete the required arch and its stone units. 97.— Pig. 2.— Solution 4. — Let A B be the span, and O D the altitude. Construct on altitude C D the equilateral triangle DEC, make C F=C A, etc., complete similar to Fig. 1. 98. — Pig. 4. — Problem.— To construct an elliptic arch when span and altitude are given. Solution. — Let A B be the given span, C D the altitude. Produce A B, and with radius C D' — C D^the given altitude describe semicircle J D'A. Divide J A and span A B similarly into the same number of equal parts, and erect at all division points perpendiculars. With the T square make CD =C' D', 2 E' and 4 E"=2 E, 1 G' and 5 G"=l G, etc.. and connect points B G" E" D E' G' A by a free- hand line, and complete the required elliptic arch. 99. — Figs. 3 and 5. — Problem.— To construct ascend- ing arches when span and altitude are given. Solution 1. — Let A B be the given span and C B the altitude; draw C A, the ascending line. Make B D (the produced span)= B C, and bisect A D by the perpendicular EG; E is center, E A the radius to quadrant AG. F C is parallel to A B, F the center, F G the radius to quadrant G C. A G C B is the required arch. Complete and add the stone units as in previous constructions. 100. — Fig. 5. — Solution^.. — Let AB be the given span, and B D the altitude. Draw D A, the ascending line, and bisect A B by the perpendicular F C: bisect angle F E A by G J, and with J as cen- ter, J A as radius, describe arc A P. Draw F J, then D H parallel to A B, and with center H, radius H D describe arc P D; A F D B is the required ascending arch. Complete and add stone units as in previous constructions. Flff. 1. '<'^' L J-J/>v "iC^ \ I// Fi^-- 2- Plate 22. Fig- 3. ff:>i — Fig: 4. .Fig. 5, AY,--^, U—-. o_ 1 L-" 1 ~~~^ Hanslein'a CoiWfrucMoe Draifffnt^. IONIC SPIRALS. 101. — Figs. 1 and 1 A.— Problenii— To construct an Ionic spiral when the altitude is given. Solution. — Let A B be the given altitude. Divide A B into 16 equal parts. The center of the spiral eye is situated in the 9th part from B, and its radius = ^ of A B. Pig. 1 A.—Bemarlc.—To explain division and sub- division, the eye of the spiral in double size is represented in Fig. 1 A. It is advisable to exe- cute Fig. 1, Plate 23 and Fig. 1, Plate 24 in as large a scale as possible, to facilitate an ac- curate division and subdivision. Draw vertical and horizontal diameters of the spiral e\c and upon these as diagonals the square D H G F. Draw then its diameters 1, C, 3 and 2. C. 4. Trisect the semi- diameters 1— C, 2 — C, etc., by the points 5 — 9.6 — 10, etc. Draw horizontal lines to the left, through points 1—2, 5—6, 9—10; to the right through points 11—12. 7—8 and 3—4; then vertical lines downward through 2—3, 6—7, 10—11; upward through 12—9, 8-5 and 4—1. The spiral is composed of a series of Quadrants, whose vertices are the points, beginning with 12 down in order to 1, the last. The limits of each quadrant are determined by the vertical and the horizontal that start from the point in question as per above. Thus the verte.K of the first quadrant is point 12, its limits the lines 12— M and 12— I, its radius the distance of 12 from D. the beginning of the spiral. The second quadrant begins at I, its vertex is point ll, its limits the lines 11— I and 11— K, etc. There are three sets of four quadrants each. The second curve is similar to the first and can easily be understood if one remembers that for every point of the primary curve there is a corresponding point of the secondary curve. The set of centers of the secondary curve is found by trisecting the distances 12-C, 11— C, 10-C, 9— C, 8—12, 9—11, etc., and using the point nearest the points 12, 11, 10, 9, etc., in the same manner as in the primary curve. ELLIPTIC ASCENDING ARCHES. 108.— Fig. 2.— Problem.— To construct an elliptic as- cending arch when span and altitude are given. Solution. — Let A B be the span and B C the alti- tude. Draw C A, the ascending line, and describe on A B as diameter, a semicircle, A NO B. Divide the diameter into any number of equal parts (6) and erect in each division perpendiculars, at which we make 2' N' = 2 N, 4' P = 4P and N' R' = N R and connect R' O' P' N', etc., by a free-hand line, which is the required arch. Bemarh. — This curve is also applied at the base of the Ionic column, as Fig. 6, Plate 9. 103.— Fig. 3.— Problem.- Tb construct an elliptic as- cending arch when S2}an, its ascending and mean altitudes are given. Solution. — Let A B be the given span, B C the ascending and E F the mean altitude. With the mean altitude E P ^ E' F' describe the quadrant F' H A E'; divide radius E' A in 3 and subdivide the last 3d into 3 equal parts. Divide the span into the same number of proportional parts and erect perpendiculars. Transfer the altitudes of F' H J, etc., to the perpendicular A D, and draw lines parallel with the ascending line A C, to obtain the points of intersection J' J", H' H", F, etc., which points, connected by a free-hand line, will give the required arch, C J" H" F H' J' A. Fig. 1. Fig. 1. A Plate 23. Fig. 2. /ill 1^- I.H-T 1 l' 1 1 ^ 1 1 1 1 1 III 4 5 K HitTuKcin's CorwIiKiI.U'e Drniuino. .— FiGS.l and lA.— Solution.— This spiral differs from that of the preceding plate, in that its altitude A B is divided in 14 equal parts. The center of the spiral eye is the eighth point from B, and as in the previous case, its radius is equal to one of the divisions. The points 12, 11, 10, 9, etc., are obtained by constructing the square D H E F (see Fig. lA), then draw its diameters 1 C 3 and 2 C 4. The semi-diameters are divided into three parts as follows: They are first bisected, and the part nearer C is again bisected. Now we are ready to start as in problem 101. The centers of the second curve, analogous to those of the primary are determined thus. Each innermost division of the semi-diameter is bisected for the centers of the first set of quadrants. Likewise the second division for the second set of four centers. For the remain- ing four divide the outer larger part of the semi- diameter into four equal parts and use the points nearest the ends of the diameter, i. e. 1, 2, 3 and 4. 105.— Fig. 3.— Problem.— To construct a spiral with semi-circles when the spiral "eye" is given. Solution. — Let C, a small circle, be the given spiral eye. Draw and produce a horizontal diam- eter, M A B N. With A as center, A B as radius, describe the semi-circle BO; C as center, C O as radius, semi-circle OP; A as center, A P as radius, semi-circle P R, etc. Curve B O P R, etc., is the required spiral. 106.— Fig. 3.— Problem. a given triangle. -To construct the evolute of Solution. — Let A B C be the given triangle. Pro- duce C B, B A and AC; B is the center, B A the radius to arc A N; C the center, C N the radius to arc N O; A the center, radius A O to arc O P, etc., etc. Curve A N O P, etc., is the required e volute. Plate 24. HansUin's Constructive DrawinQ. CAM LINKS — AHCUIMEDEAN SPIRALS. Definition.'— An archimedean spiral is a curve in a plane generated by a point whjse distance from a centre of rotation increases uniformly. Cams are arrangements in mechanics by which a rotary motion is converted into a reciprocating action, they are con- structed by archimedean spirals. Remark. — The following curves, used principally in mechan- ics and architecture, should be executed by free-hand lines before the student attempts to use a curve rule. 107. — Fig 1.— Problem.— To construct a cam-line of lU revolutions wlien the distance C C between revolutions Is given. Solution.— \fith 8 equal parts, 6 of which are equal to the given distance C C, describe the circle 8 A B D E F, which is divided into 6 equal parts by diameters. Describe circles with C as center, radius C J, to intersect diameter B F in B'; with radius C 2 to intersect D 8 in D'; C 3 to intersect E A in E', etc. ; connect points G B' D' E' F' 5 C H D by a free-hand line, to complete the required cam-line. 108.— Fig. S.— Problem.— To construct a heart-shaped cam wlien the altitude is given. Bcmar/c.— Heart-shaped cams are made to convert half of a revolution into forward motion, the other half of the revolution into backward motion. ( Piston-rods for pumps, etc.) Solution —Let G 8 be the given altitude, which is divided into 8 equal parts and is the radius, C the center of the circle, divided by diameters into 16 equal parts. With center C, radius C 1, describe circle to intersect radii C A and C G in A and A', with C 8 as radius to cut radii B C and J C in B' and B", with C 3 to cut radii G C and K C in C and O", etc. Connect CA' B' C D'l EHG" B" A" C by a free-hand line and complete the required heart-shaped cim. 109. — Fig. 3.— Problem.— To construct a cam in 4 equal divisions, to raise a lever in the first a of its revolution, equal to the altitude B D, to remain stationary the second a, to descend its first position the thii-d %i. and remain stationary the last a of its revolution. SoluMon.- Let B D be the given altitude, B A an arbitrary tance from the hub, and C the center of the cam. Describe with C D, center C, the circle D H D' 4 and divide it into quadrants, two opposite ones Into 1 equal parts again, by diameters N 4 =• B D " the given altitudeis also divided iuto4equal parts, 1, 2, 3 and 4, and with radius C 1 draw arcs 1 G', G, with radius C 3, 3 F' F, wit radius C 3, 3 E' E; connect N G' F' E' D' and the symmetric points B G F E H by a free-hand line and com- plete the required cam. 110.— FiQ. 4.— Problem. To construct a cam in three equal divisions, which in one revolution shall lift a lever — A 4 in theffrst 3d, shall remain stationary the second 3d, and shall rise again the third 3d an altitude — 4 B and make a sudden escape at B, to renew its motion in the second re«oiu(ton . Solution.— het the two inner circles be shaft and hub circum- ferences. A 4 the altitude of the first incline, 4 B the altitude of the second incline (the third division). Remark. — This construction, in applying the principles of Figs. 1, 8 and 3, will not present any difficulty to the student, and ( 1 now be solved without the assistance of a teacher. CONIC SECTIONS. ELLIPSE, PARABOLA AND HYPERBOLA. 111. — Fig. 5, — Three curves, which we obtain by sectional planes through a circular cone and cylinder, are of the greatest im- portance in technical work; the ellipse, parabolaand hyperbola. A sectional plane through the cylinder or circular cone in an oblique direction, as U V or M N, respectively, creates the ellipse. A sectional plane S T, parallel to the side B of the circular cone, creates the parabola. A sectional plane Q R, parallel to the axis of the circular cone, creates the hyperbola. Definition.— An ellipse is a closed curve; the sum of the dis- tances of each point in this curve from two iixed points within, called foci, is equal to the long axis. The ellipse has two axes, the major and minor, bisecting each other perpendicularly and dividing the ellipse as well as its surface into two symmetric parts. 112.— Fig. 6.— Problem.— I'o construct an ellipse when major axis (Iransversant) and minor axis (conjugant) are given. Solution.— Let A B be the major, C D the minor axis. With a radius H A B = A M and center C draw arc and intersections with A B, points F and F', the foci; divide F M arbitrarily into parts, increasing in length towards M, and with F and F' as centers, B 4 as radius, describe arcs E G and E' G'; with F and F' as centers, A 4 as radius, draw intersections at E and G and at E' and 6'. Points E E' G G' are situated at the circumference of the ellipse. Operate with points 3, 2 and 1 in the same manner, and we obtain by each operation 4 points, which lie at the circumference of the ellipse, as with point 2, e g,, by which we locate points H J H' J'. Connecting these points by a free hand line, we obtain C E H A J G, etc., the required ellipse. Fig. 1. Fig. 2. Plate 25. ■Fig. 3. Fig. 4. Han&tcirCs Confiirwtfue I>rawinQ% H3.— FiQ.l — Problem.— To construct a tangent to an ellipie when the point of contact is given. Solution.— Let A B D be the ellipse and G the point of con- tact. Describe from G as center, withradius G F, the arc F N, and draw and produce line F' G, intersecting arc F N in N; bi- sect angle N Q F by IJ, which is the required tangent. Bcmark— In elliptic arches, executed in cut stone, the joints are perpendiculars (as P 6) to tangents, having the unit divisions as points of contact. lU.— Fig 1.— Problem.— F/'om an exterior point to construct a ta7igent to an ellipae. Solution.— liet H hi the given exterior point. With H as center, H F' as radius, describe arc F' O; with A B as radius, and F as center, intersect arc F' O in O. Bisect arc F' O by L H, which Is the required tangent. IIB.— Fig. 3, — Problem.— To constnict an ellipse when both axes are given. (Practical solution.) Solution 1. — Let A B and C D be the given axes. Find the foci (118) and place in F, F' and C pins, around which tie a linen thread to form the triangle F C F '. Take away the pin at C and place the pencil point in the triangle, by stretching the thread gently and forming a vertex of the triangle; draw the curve, which will be the required ellipse. Solution 2 —A B and C D are the given axes. Take O P, a straight edge or a slip of paper, at which make A' M' = A M = 14A.B and A' C = C M = V4 C D. Guide the straight edge to have point C follow the major axis, and M' the minor axis, then will point A' describe the circumference of the required ellipse. Locate the position of point A' during this operation by pencil marks, which, connected, will give the ellipse. Bemo»/t.— Place In points C and M' pins, in point A' a pencil point, and let these pins slide in grooves In the place of the axes; we have an instrument called a trammel or ellipsograph, with which we are able to draw any ellipse by arranging points A' C and M' in the required proportions. 116.— Fig. 3.— Problem.— To construct an ellipse by intersecting lines. Solution.- Let A B and C D be the given axis, and construct with these lines the rectangle E F G H; divide A B and E G into the same number of equal parts and number as in the diagram* Draw lines D 1 P, D 8 O and D 3 N, intersecting the lines C 1, CS and G 3 at P, O and N, etc., which points, connected by a free- hand line, will be the required ellipse. 117.— Fig. *.— Problem.— To construct an elliptic curve in an oblique parallelogram. Solution.— Let B F G H be the parallelogram. Draw axes A B and C D bisecting opposite sides, and divide M and B C into the same number of equal parts; proceed as in the previous construction and draw C P O N A, etc., the required ellipse. 118.— FiQ. 6 A.— Problem.— To construct an ellipse by intersec- tions of lines. Solution.— With A B and C D, the given axes, construct the rectangle B F H Q; divide E O and A E in the same number of equal parts (4) and number as shown In the diagram. Draw lines 1 A, 2 3, 3 2, and C 1, and connect their intersections T S R, etc., by a free-hand line to complete C T 8 R A, etc., the re- quired ellipse. 119. — FiG.5 B.— Problem.— To construct an ellipse by its tangents. Solution.— Dt&w and divide C B into any number of equal parts (4): 1, 2, :i and 4, through which parallel with D draw P P', O O' and L L'; draw also E 1 1, E 2 K and ESN and lines LN,OKandPI, which are the tangents to the req uired ellipse. Draw the ellipse by a free-hand Une. 120.— Fig. ().— Problem.— To construct an ellipse by the differ- ences of two circles. Solution, — Let B A and C D be the given axes. Describe with B A and C D as diameters concentric circles with center M. Divide both circles into 12 equal parts by the diameters 10, 4—11, B— 1, 7—2, 8 and 8, 9. Draw lines 7, 5—8, 4—10, 2 and 11, 1, and from the intersection points E F G and H the perpendiculars to 10, 2—11, 1—7, 5 and 8, 4, which will give points N O A P K D, etc., at the circumference of the required ellipse. Fig. 1 Fig. 2. Plate '26 Fig 3. \ ^^^\ ""a5{i Fig. 4. Fig 5. Fig 5. Fig. 6. UanaUin'e Constrictive DrawinQ. PASABOLl. 131.— Fig. 1.— Problem.— To construct a parabola when the axis and the base are gioen. Definition.— The parabola is a curve In which the distance of any point from an outside right line (direotrixj is equal to the distance of this point from a fixed point within, called focus. A line bisected perpendicularly by the asis at its terminus and Intersecting the curve is called the base, and a parallel with it, through the focus, the parameter of the parabola. Solution.- Let AP be the axis and LU the given base. Bisect LP = Vi the base L K in J, and draw J A. .In J erect a perpen- dicular to J A, J R intersecting the produced axis in R; trans- fer P R to left and right of point A, to obtain point F, the focus, and point O, through which draw M N, the directrix, perpen- dicular to the axis O P. Divide A P into arbitrary parts, 1, 2, 8, 4, etc., in which erect perpendiculars, and with F as center, O 1 as radius, cut the perpendicular 1 in B and B'; with O 2 as ra- dius, the same center, cut the perpendicular 3 in C and C ; with O 3 as radius cut perpendicular 3 in D and D', etc., and connect the obtained points L B' B' A B B K by a free-hand line, which is the required parabola. 132.— Fig. 3,— Problem.— To construct a tangent to a parabola when the point of contact is given. Solution.- Let L RK be the given parabola, O P the axis, M N the directrix, and A the point of contact. With A as center, A F as radius, draw arc F B and A B perpendicular to M N. Bi- sect arc F B by line S G, which is the required tangent. Problem.— To construct a tangent to a parabola from an ex- terior point, R. Soiution.- With E as center, and B F as radius, draw arc F D and erect at D a perpendicular to M N, Intersecting the para- bola in H, the point of contact ; or bisect arc D F by line T E, which is the required tangent. 123.— Fig. 3.— Problem.— I'o construct a parabola when two gym' metric tangents are given. Solution.— het B E = A E be the given tangents. Divide E B and A E into equal parts and number as shown in the diagram. Draw lines 7-7, 6-6, 6-5, 4-4, etc., which are the tangents of the parabola. A free-hand curve tangential to these tangents is the required parabola. 124.— Fig. i.—Probltm.— To construct a parabola wTien the axis and the base are given or the rectangle drawn with these lines. Solution.— het A B 6 J be the given rectangle. Divide H 6 J = D 6 and B 6 into 6 equal parts, respectively ; number as in the diagram, and draw parallel to the axis D O lines through 1, 3, 3, 4, 5. Draw also lines 5D, 4D,3D,3DandlD, intersecting with the horizontals in points I H G E F D, etc., which points, con- nected by a free-hand line, furnish the required parabola. 125.— Fig. 5.— Problem.— To construct a parabola practically when base, O P, and axis, A B, are given. Solution.— Locate the focus F and the directrix M N and place a straight edge firmly coinciding with it. Fasten a thread to a pin placed in F and pass it around a pin in A to a point D of the set square, when its side G D coincides with axis A B. Remove the pin in A and hold the pencil to stretch the thread gently, touching C D constantly, shift the set square to the left. The pencil point will describe the required parabola on tha drawing paper. 136.— Fig. 6.— Problem.— To construct a Oothic "rch by para- bolas. Solution.- Let A B be the span and F E the altitude of the arch. Construct the rectangle C D B A, divide C D into 8 and E F into i equal parts and number as the diagram. Draw lines 1 A, 2 A, 3 A and parallel to span III', J 2 J', and H 3 H'. The points of intersection, A IJ H B H' J' I' B, connected by a free- hand line, complete the arch. Fig. 1. Fig. 2. Plate 27. Fig. 3. Fig. 4. Fig 5. Fig. 6. ■^ ^^^1 ^-^ ^^:^_ rtrir-i BaneUin'a Comtruetiee DrawliUr- 127. — Fig. 1.— Problem. — To construct hyperbolas when the vertices and foci are given. Definition. — The hyperbolas are curves; the dif- ference of distances of each point to the foci is equal to an invariable line, the axis. Solution. -Place on line M N, A and B the ver- tices, and F and F' the foci equidistant from O. From F' towards M mark arbitrary divisions and number as in diagram. With radius B 1, cent r F, — radius A 1 and center F' draw intersecting arcs at C and C; radius B 2, center F and radius A 3 and center P' draw intersecting arcs at D' and D, etc. Connect G' E' D' C A C D P G by a free- hand line, to complete the required hyperbola. To obtain the second curve, operate symmetrically. 128. — Pig. 2.— Problem. — To construct a tangent to a hyperbola idhen point of contact, P, is given. Solution. — Draw line P F, and with radius P F' and center P the arc P' D. Bisect F' D by the line T U, which is the required tangent to the hy- perbola. Bemark. — The stone joints in hyperbolical arches are the perpendiculars to tangents at the point of contact. Problem. — From an exterior point, R, to construct a tangent to the hyperbola. Solution. — With R as center and radius R P draw arc P N; with P' as center and radius A B cut arc P N in N and bisect P N by S R, which is the required tangent to the hyperbola. 139. — Pig. 3.— Problem — To construct hyperbolas when axis A B «s given; to find foci and draw the asymptotes. Asymptotes are right lines to which the branches of the hyperbolas approach when produced, but do not touch. Solution. — Construct the square EDCG with C D = A B, which the axis divides into two equal rectangles. Draw and produce the diagonals M N and O P, which are the required asymptotes. With O as center, O G as radius, draw arcs G P' and C F. With F and P', the required foci, draw the hyperbolas, as in Fig. 1. EVOLUTE. 180.— Pig. 4.— Problem.— To construct an evolute at a given circle. Definition. — An evolute is a curve made by the end of a string unwinding from a cylinder. Solution. — Let C be the given circle ( the section of a cylinder). Divide the circumference into a number of equal parts (12) and draw the diameters and tangents 1 A', 3 A", 3 A'", 4 A*, etc. With center 1 and radius 1 A describe arc A A'; center 2, radius 2 A', the arc A' A"; center 3, radius 3 A", the arc A" A'", etc.; curve A, A', A", A'" is the required evolute. Fig. 1. Plate 28 Fiff. 2. Fig. 3. Fig. 4. HavMeWt C&nstructiv€ Drawing. GEAR LINES— CYCLOID. 131. — Fig. 1. — Problem. — To construct a cycloid when the generating point A is given at the cir- cumference of the circle. Definition.— k. cycloid is a curve generated by a point at the circumference of a circle, making one revolution in rolling on a straight line. The curve generated, when the circle rolls on the outside cir- cumference of another circle, is the epicycloid, and when the circle rolls on the inside circumference of another circle, the hypocycloid. Solution 1. — Let C be the rolling circle, tangent A B its rectified circumference and A the generat- ing point. Divide the circle C and line A B into the same number of equal parts (12) and number as in diagram. Pass horizontals through points 1, 2, . 3, etc., of the rolling circle and erect perpendicu- lars at A B in points 1, 2, 3, etc. With points C, C", C^, C-" as centers, C A as a radius, describe circles 1 A', 2 A", 3 A'", 4 A*, etc., which points connected give the required cycloid. 132. — Fig. 2.— Solution 2.— Follow the operations of the previous construction. Draw the circle C 6, also chords 6 1, 6 H, 6 G, G F, 6 E and their symmetric chords. Parallel to 6 E draw E' 7 K, to 6 F, F' 8 L', to 6 C, G' 9 M', to 6 H, H' 10 N' and to 6 I, I' 11 O'. 11 is the center, radius 11 B for arc B I', O' the center, radius O' I' to arc I' H', N' the center, radius N' H' for arc H' G', M' the center, radius M' G' to arc G' P', L' the center, radius L' F' to arc F' E', and K E' the radius to arc E' D E. Com- plete the construction symmetrically to the left of axis D K. The curve of B I' H' G', etc., is the required cycloid. When a cycloidal arch is executed in stone, the radii of the pertaining arcs are the joints of the units. 133. — Fig. 3. — Problem.— To construct a cycloid when the point generating the curve is situated at a greater radius than that of the rolling circle. Solution. — Let C G be the rolling circle, G 12 its rectified circumference and A the generating point. Describe with C A from C a concentric circle and proceed in this construction as in Fig. 1. Pass horizontals through the divisions of the greater circle aud describe with radius C A and centers C, OS C\ etc., the circles B A', C A", D A'", etc. The curve passing through points A, A', A", A'", etc., is the required cycloid. Plate 29. Fig. 2. V" Fiff, 3. Han&iein's Conslructiot Drawing. GEAR LINES — EPICYCLOID AND HYPOCYCLOID. 134. — Fig. 1. — Problem. — To construct an epicycloid when the relation of the rolling to the stationary circle is 1 : 2. Solution. — Let A B and 6 A be the diametei-s ol' tbe given circles, having the proportion of 2 : 1, respectively. Divide the rolling circle into any number of equal parts (12), and as circumferences are proportional to diameters, the circumference of 6 A = the semi-circumference B A contains 12 of the same equal parts. With center C draw circles passing through points 1, 2, 3, 4, 5, 6 and D, and also the radii D'a, D"b, D"'c, etc. .D', D", D'" are the centers and radius D A to arcs aA', bA", cA'", etc. Connect A, A', A", A'", etc., by a curve, which is the required epicycloic. 135. — FlG.2. — Problem. — To construct a hypocycloid- Solution. — Let C be the circle, point A the gener- ating point rolling in circle E. Relation of circles 1 : 8. Make an equal division in both circles (A b = A 1) and draw radii A C, b C, c C", etc. C, C, C", C", etc., are the centers and C A the radius to arcs b A', c A", d A'", etc. Connect A, A', A", A'" by a curve, which is the required hypocycloid. 136. — Fig. 3. — Problem, — To construct a. hypocycloid fvhen the relation of the circles is as 1 : 2. Solution. — Treating this construction as the pre- vious one, we shall obtain a right line A B as the required hypocycloid. This construction is the fundamental principle of the planet wheel, applied to convert directly a rotation into a reciprocating movement (pump- piston). 137. — Pig. 1. — Problem. — To constntct an epitrochoid. It will not be difficult to execute this curve. See Fig. 2, Plate 29. 138. — Fig. 2. — Problem. — Tv construct a hypotro- choid. Solution. — Let C F be the rolling circle, A the generating point and D J B the circumference on which circle C rolls. Proceeding as in Figs. 2 and 3, we obtain the curve A, A', A", A'", etc., which is the required hypotrochoid. APPLICATIONS TO ARCHITECTURE. 139.— Fig. 1.— Problem.— ro construct a design for an ornamented Gothic arch in stone. This construction is based on principles ex- plained and described in the previous part of this volume, and its solution should not present any serious difficulties to the student. Bemarlc. — To obtain an accurate result, it is ad- visable to make the equilateral triangle, the fun- damental figure of this arch, not less than 8 inches a side. APPLICATIONS TO MECHANICS. 140.— Fig. 1.— Preblem.— To construe* a pair of spur-wheels, their relation tohel : 2. To solve this problem we require the construc- tion of two epicycloids and two hypocycloids to the "flanks" of the teeth, and it is advisable to enlist the advice of a teacher, to execute this im- portant construction correctly- Pkess of Max Stern's Sons Chicago ./ LIBRARY OF CONGRESS QDD5S1754TT